Additional Exercises with Solutions for:

A First Course in Applied StatisticsWith applications in Biology, Business, and the Social Sciences

Second edition

Megan J. Clark John A. Randal Alysha O. Nickerson

March 21, 2010

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This document is intended as support for the second edition ofClark and Randal, A First Course in Applied Statistics, Pearson,2010.The exercises are from the first edition, and these are not repeatedin the second. The solutions for these were largely produced byAlysha Nickerson. The authors are particularly grateful to Alyshafor her dedication to this project.Solutions are given for all end-of-chapter exercises in the book,and we include slightly expanded duplicates of the solutions givenin Appendix D of the first edition.All graphs were produced by the statistical software R (availablefor free at http://www.r-project.org/).Please report any errors or omissions to john.randal@vuw.ac.nz.

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Chapter 2

Exercises

The first set of questions gives you practice classifying variables.

2.1 The following table is the first two lines of a large data set of information collected on 200 childrenreferred to Psychological Services for assessment for possible attention deficit hyperactivity disorder.

Age Gender Maximum Activity level Number of times1 = M concentration 1 = low broke away2 = F span on 2 = medium from task

task (min) 3 = high

9 1 2.5 3 313 2 5.8 2 12...

......

......

Classify the five variables according to whether they are quantitative or categorical. If they arequantitative, further classify as discrete or continuous.

2.2 A graduate student in ecology is collecting data on cicadas. The following is the beginning of herdata set:

Maturity Location Singing time Body weight Number in Time of dayjuvenile/adult (min) (g) quadrat

J B 0.8 2.8 7 dawnA B 4.5 3.3 8 mid am...

......

......

...

Classify these six variables according to whether they are quantitative or categorical. If they arequantitative, further classify them as discrete or continuous.

2.3 Classify the following variables as quantitative or categorical. If they are quantitative, classifyfurther as discrete or continuous.

(a) Size of student loan.(b) Maori population of New Zealand in 2050.(c) Maximum flow (cubic metres per second) of the Hutt River.(d) Residence of people who commute to Wellington to work.(e) Weight of weta faeces.(f) Amount of leaf browse in New Zealand mistletoe in a region (high, medium or low).(g) Habitat of spotted skink.(h) Concentration of mercury (mg/m3) in Lake Rotorua.(i) Age when sent to prison.(j) Reaction time to auditory stimulus.

The next set of exercises give practice drawing stemplots and dotplots.

2.4 In a study of sixty individuals with brain damage in which word recall is apparently blocked byan intruder word, the number of inappropriate words used instead of a correct response was notedduring an hour of questioning, and recorded as the number of intruder words per 10-minute interval.When the data are organised according to the age at which brain injury occurred (and those withunknown age of injury deleted) the data are as follows (ordered from lowest to highest)

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Injured before age 15 Injured after age 15

7.1 7.2 7.2 7.9 8.1 8.4 8.7 9.0 9.6 9.88.1 8.3 8.4 8.6 9.0 10.1 10.5 11.1 11.2 12.49.0 9.8 9.9 10.1 10.1 14.3 14.8 15.2 15.6 15.610.5 11.0 12.5 14.3 15.4 15.7 15.7 16.0 16.2 16.416.0 16.1 16.8 17.0 17.1 17.5 17.7

17.7

(a) Draw back-to-back stemplots of this data.(b) Draw side-by-side dotplots of the data for those injured before the age of 15 and those injured

after age 15.(c) Comment on any features of these plots. Which seems more informative? Why?

2.5 The owner of a Subaru Legacy monitors the petrol consumption of his car. The following data arethe number of kilometres driven per litre (km/L) of petrol consumed for 30 tanks of gas:

11.6 10.7 11.5 10.1 10.5 9.6 10.9 11.0 10.6 8.59.7 9.3 8.6 8.2 8.7 10.6 10.8 9.4 8.5 10.19.6 11.1 11.6 8.5 11.3 11.6 9.3 10.3 10.2 10.5

Draw a split-stem stemplot of these data. Does this plot indicate any interesting patterns in thecar’s fuel consumption?

2.6 Eggs from the Stephen’s Island tuatara (Sphenodon punctatus) were collected in November 1998,and incubated in the lab at 18, 21 or 22 degrees Celsius. After hatching, the young tuatara hadtheir development monitored. This data set is a random sample of ten from each incubationtemperature, although the incubation temperatures have not been noted. The measurement isSVL = snout-vent length (mm) at age 10 months.

80 82 81 82 73 85 81 83 80 8291 84 83 90 83 87 89 85 82 8481 83 84 87 76 84 88 83 91 84

Draw up a dotplot of this data. Is the effect of the three different incubation temperatures showingin the plot? How?(Data provided by Nicola Nelson, from Nelson, Nicola J., Michael B. Thompson, Shirley Pledger,Susan N. Keall and Charles H. Daugherty (2004). Egg mass determines hatchling size, and in-cubation temperature influences post-hatching growth of tuatara Sphenodon punctatus. Journal ofZoology, 263, 1–11.)

2.7 A study of a wetland site to investigate the rate of regrowth of vegetation after a catastrophic galehas as part of the study the measurement of leaf lengths of a particular shrub at a date 12 monthslater. Twenty-eight plants were randomly selected and the following is the length of the largestleaf per plant (cm):

6.5 8.3 5.5 7.8 10.1 9.7 8.1 7.2 9.6 9.911.4 8.4 14.2 6.4 8.8 12.7 14.5 6.4 10.3 8.313.3 10.7 7.8 11.4 8.7 6.4 6.7 13.1

Draw a dotplot of this data. Summarise the information in the dotplot. Now draw a stemplot.Which do you prefer and why?

2.8 A member of the rush family, Rostkovia magellanica is found only in a few locations such as theCarrick Range in Central Otago and Takahe Valley in Fiordland. Lengths of 34 fully grown plantsin samples taken in these locations were as follows (cm):

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Carrick Range 12.7 14.4 13.0 12.6 12.8 13.3 12.4 13.5 13.614.3 14.3 13.9 13.9 13.2 13.3 12.7 12.9 13.712.9 13.0

Takahe Valley 12.6 12.7 13.6 13.1 12.3 12.8 11.5 12.3 13.012.5 12.7 13.3 12.5 13.1

Draw a back-to-back split-stem stemplot of these data. What are the main features of the data?

2.9 The following data are alcohol consumption scores for n = 19 eighteen-year-old males. The amountshave been scaled to account for differences in body weight to make them directly comparable: 16,33, 17, 8, 19, 22, 22, 18, 25, 32, 14, 18, 17, 18, 10, 9, 22, 14, 14. Draw a split-stem stemplot torepresent this data. Why is a split-stem sensible in this situation? Use this plot to find a sensiblerepresentative value to describe 18-year-old males’ alcohol consumption in a single value.

These exercises involve frequency tables, and histograms. They also provide further practice with variableclassification, and stemplots.

2.10 Thirty-six soccer players having their fitness assessed had to do as many sit-ups as possible in12 seconds. Each player had to do this task five times in 4 hours and their average number ofsit-ups/12 seconds was recorded. The data are as follows:

3.7 4.8 2.0 4.0 1.6 2.7 4.7 4.8 4.64.1 3.8 1.4 2.8 4.5 1.8 4.7 4.4 3.84.5 2.3 1.5 4.7 1.6 4.2 4.4 4.4 1.14.0 3.5 4.0 1.2 1.1 4.9 1.3 1.3 1.4

(a) Draw up a frequency table of the data using the following class intervals: 1.0 -(1.5), 1.5-(2.0),. . .,4.5-(5.0).

(b) Use the table in (a) to draw a histogram of the data and draw a frequency polygon on thehistogram.

(c) Summarise the main features of this data.(d) Draw a split-stem stemplot of the data. How does this compare to the frequency table in (a)?(e) The data above is part of a wider set of data collected on the players. The following are the

first 2 lines of that data set:

Gender Pulse rate Ave. situps Body weight Teamfemale=0 beats per min (/12sec) (kg)male=1

0 67 4.5 72.4 D1 52 1.2 59.2 B...

......

......

Classify the 5 variables according to whether they are quantitative or categorical. If they arequantitative, further classify as discrete or continuous.

2.11 Draw a histogram of the following monthly returns (measured in %) from five years’ share pricedata for Telecom New Zealand. Comment on any features.

−3.9 −3.7 7.0 4.2 13.1 −6.6 −11.8 −2.3 −7.0 3.1 −10.3 2.70.2 7.6 7.0 −3.3 −10.0 −0.9 −0.5 −8.8 −10.8 −4.1 −4.8 −9.7

−10.2 7.6 4.5 −1.1 −5.2 11.9 1.9 0.6 −8.5 −0.5 7.4 −5.9−2.6 −9.7 2.5 1.2 2.9 8.6 0.1 4.8 4.5 −5.4 −11.0 5.4−1.9 0.7 −5.9 4.1 5.5 4.7 −12.0 4.5 4.5 −0.6 3.6 −1.0

2.12 In the study of sixty individuals with brain damage mentioned in Exercise 2.4 in which word recallis apparently blocked by an intruder word, the number of inappropriate words used instead of acorrect response was noted during an hour of questioning, and recorded as the number of intruderwords per 10-minute interval. The results are shown below:

5

Words 7-(8) 8-(9) 9-(10) 10-(11) 11-(12) 12-(13) 13-(14) 14-(15) 15-(16) 16-(17) 17-(18)

Freq 4 8 9 6 4 2 0 3 6 8 10

Draw a histogram for the data and on this histogram draw a frequency polygon.

2.13 The shrub Pimelea concinna is found mainly on the upper reaches of the Wairau and Hurunuirivers. The following table illustrates the numbers of Pimelea concinna found in each of thirty10m2 sites on the upper reaches of the Wairau river:

Plants per site 0-(5) 5-(10) 10-(15) 15-(20) 20-(25) 25-(30) 30-(35) TotalFrequency 8 5 6 4 3 3 1 30

(a) Draw a frequency histogram of this data.(b) As accurately as possible draw the frequency polygon of the data above.(c) If the data are to be compared with data from 52 sites on the Hurunui river which would you

use on the vertical axis; frequency or relative frequency? Why?(d) Why is a histogram not a particularly suitable form of representation of this data?

2.14 The following histogram illustrates the numbers of Rostkovia magellanica found in each of 33 siteson the Carrick Range in Central Otago.

0 20 40 60 80

freq

uenc

y

number of plants at site

4

2

10

8

6

(a) Find the number of sites in each of the groups and record in a frequency table.(b) As accurately as possible, draw the frequency polygon on your histogram.(c) If these data are to be compared with data from 45 sites from Takahe Valley in Fiordland,

which would you use, frequency or relative frequency on the vertical axis? Why?(d) What type of data is plants per site?

2.15 The following data represent the body lengths of a sample of 260 paua seized by Fisheries staff inJanuary. (Note that paua must be 12.5 cm in length before they can legally be taken.)

Length of paua (cm) Frequency

7.5 - (10.0) 3510.0 - (12.5) 7012.5 - (15.0) 8015.0 - (17.5) 2017.5 - (20.0) 1320.0 - (22.5) 3022.5 - (25.0) 12

(a) Why is a histogram more appropriate than a bar chart for these data?(b) Draw a histogram for this data showing relative frequency on the vertical axis. Describe this

histogram in a single sentence. What does it suggest about the paua stock?

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2.16 A random sample of 500 families from South Auckland were scored on a check list that is used togauge the availability of extended family support, and the data presented in the following table:

Availability of family Availability of familysupport score Frequency support score Frequency

0.0 - (1.0) 22 6.0 - (7.0) 651.0 - (2.0) 35 7.0 - (8.0) 802.0 - (3.0) 50 8.0 - (9.0) 433.0 - (4.0) 70 9.0 - (10.0) 354.0 - (5.0) 45 10.0 - (11.0) 205.0 - (6.0) 35 11.0 - (12.0) 0

(a) Present this data in a histogram. Why is it an appropriate choice of graph to represent thisdata?

(b) How many modes does your histogram have? What might this mean about the sample?(c) Later investigation classified 210 of the families as dysfunctional (using a standard assessment

scale) and the remainder as functional. Is this consistent with the graph? Give reasons foryour answer.

(d) Random samples of 15 families from each of these two groups yielded the following scores:

Dysfunctional Functional

4.6 5.5 3.1 10.5 6.4 9.0 9.2 9.8 7.5 6.80.5 9.3 4.8 3.3 6.8 10.9 6.0 8.1 7.8 8.93.9 1.8 4.9 5.7 10.9 10.5 10.2 8.7 9.4

Draw back-to-back stemplots of these data.(e) Summarise the main features of the data.

2.17 A producer of chocolate bars measures the weight of a particular variety at the end of the productionline. Forty randomly selected bars gave the following weights (in grams):

51.5 51.7 51.2 50.4 51.8 49.6 50.6 50.8 50.9 51.7 51.1 51.3 51.0 50.851.8 49.3 51.4 51.3 50.7 51.7 50.7 50.8 51.6 52.2 51.2 50.7 50.2 50.850.8 51.3 50.4 52.9 51.1 51.8 49.4 51.5 50.1 51.6 51.5 50.3

(a) Draw a histogram of these data.(b) Draw a split-stem stemplot of these weights. Which of these two representations do you prefer?(c) The bars have an indicated weight of 50g. Does the data suggest the producer is satisfying this

requirement?

The remaining exercises give practice drawing bar graphs.

2.18 The following information was gained from people given a diagnosis in the New Zealand mentalhealth system in 2002.

Diagnosis - disorder due to: % total patients in system

Organic disease, inc. dementias 5.2Substance abuse 17.7Schizophrenia and related 18.3Mood disorders (bipolar, etc.) 33.9Neurotic, stress and anxiety 17.2Behavioural syndromes, inc. eating disorders 1.4Adult personality disorders 3.4Mental retardation 0.9Disorder of psychological development 0.7Hyperkinetic and related 4.3

(a) Represent this data in a bar chart.

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(b) The percentages in the table add to more than 100% - explain why this might be so.

(Data from Mental Health Information National Collection.)

2.19 The following is New Zealand climate data for selected cities:

City Mean Sunshine Mean Days Mean Meanannual hours annual of annual annual

rain days per annum temperature frosts maximum minimum

Auckland 140 1904 15.7 0 28 3Gisborne 113 2173 14.1 7 33 -2Taupo 123 2015 12.0 37 30 -4Napier 92 2187 14.3 9 32 -2Wellington 117 2008 12.7 0 27 1Nelson 96 2372 12.2 38 28 -4Christchurch 85 1992 11.9 36 34 -5Dunedin 119 1645 11.1 10 29 -2

Draw a bar graph to display this information.(Data from New Zealand Yearbook, 1984.)

2.20 ICME is an international conference on mathematics education that occurs every four years. The1996 conference was held in southern Europe, the 2000 one in Asia and the 2004 conference was heldin northern Europe. There are always concerns when choosing the venue that it be as accessibleand as affordable to delegates from as many countries as possible. On the other hand, a hostcountry has to have the facilities and infrastructure to hold the conference, which typically hasover 2000 attendees. The following is data from selected countries on enrolments for ICME at atime three months prior to commencement:

VenueSouthern Europe Asia Northern Europe

Delegates from (Spain) (Japan) (Denmark)

Poland 10 4 9Italy 43 10 35China 21 114 59UK 202 88 108USA 278 222 314Brazil 149 23 27Australia 130 64 90New Zealand 16 13 14South Africa 44 13 30Mexico 20 11 15

Draw a bar graph with bars for each venue using a suitable key. Suggest some explanations for thepatterns you see there.(Data from http://www.icme-10.dk/)

2.21 The following data are from the five busiest container ports internationally in 20-foot equivalentunits (TEU, i.e. a unit can be one 20-foot container or two 10-foot containers).

Rank 2000 2001 2002

1st Hong Kong 18,098,000 Hong Kong 17,826,000 Hong Kong 19,144,0002nd Singapore 17,086,900 Singapore 15,571,100 Singapore 16,940,9003rd Pusan 7,540,387 Pusan 8,072,814 Pusan 9,453,3564th Kaohsiung 7,425,832 Kaohsiung 7,540,525 Shanghai 8,620,0005th Rotterdam 6,274,556 Shanghai 6,334,400 Kaohsiung 8,493,000

8

Note that of all these ports, Singapore and Hong Kong have by far the smallest yard areas to workin, e.g. the Rotterdam port has 280 hectares with a yearly throughput of 2.5 million TEU whileHong Kong is 122 hectares and a yearly throughput of 6.5 million. Use bar graphs to demonstratethis data as effectively as possible by stacking and, using a suitable unit of measurement.(Data provided by Yat-wah Wan from his project with Zhang, Liu, Murty and Linn.)

2.22 Wade et al worked with doctored photographs to see if they could create false memories of childhoodevents. Subjects worked at memorizing five photos of their childhood (of which one was doctored)for a week and then underwent a series of tests. For question A they had to rank their responseto: “As I remember the event, I can hear it in my mind” on a Likert scale from 1 = not at all upto 7 = as clearly as if it was happening right now. They also had to respond to B: “As I rememberthe event, I can see it in my mind” on the same scale. Responses for 20 subjects for these twoquestions were:

Question A: 1 4 3 3 2 7 1 6 4 1 3 2 5 1 35 2 3 2 4

Question B: 4 5 4 2 6 6 2 3 4 3 1 5 3 4 47 1 1 4 5

Draw side-by-side bar graphs for the responses to these two questions. What do the plots suggestabout the responses to the two questions?(Data from Wade, K.E., Garry, M., Read, J. D. & Lindsay, D.S. (2002). A Picture is Wortha Thousand Lies: Using False Photographs to Create False Childhood Memories. PsychonomicBulletin & Review 9, pp. 597–603.)

2.23 The following table gives a picture of the changes in the Maori population of particular NorthIsland regions from the 1840s to 1874. The Maori Population in the Southern North Island was:

Locality Early 1840s 1850 1857 1874

Port Nicholson 800 745 583 161Waikanae/Porirua 1020 888 483 162Otaki 1000 1379 954 491Manawatu 1000 877 709 536Rangitikei 600 259 647 407Wanganui 5800 No data 3837 2437Wairarapa 500 780 740 742

Present this information in some sort of bar graph. Summarise the main points.(Data thanks to Brad Patterson, published in McConchie, Winchester and Willis, Dynamic Welling-ton, 2000, Institute of Geography, VUW.)

2.24 The following data on student loans over $50,000 is from the Dominion Post, 21 May 2004.

Loan balance ($) Number owing

> 99,999 38280,000 - 99,999 107160,000 - 79,999 450955,000 - 59,999 252750,000 - 54,999 3709

Draw up a bar chart of this data with frequency on the vertical axis.

Solutions

2.1 Categorical: gender, activity.Quantitative continuous: concentration, age (but usually recorded as quantitative discrete).Quantitative discrete: number of times.

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2.2 Categorical: maturity, location, time of day.Quantitative continuous: singing time, body weight.Quantitative discrete: number in quadrat.

2.3 Categorical: d, f, g.Quantitative continuous: a, c, e, h, i (but usually recorded as discrete), j.Quantitative discrete: b.

2.4 (a)Under 15 Over 15

9 2 2 1 76 4 3 1 1 8 4 7 key

9 8 0 0 9 0 6 8 8 | 4 = 8.45 1 1 10 1 5

0 11 1 25 12 4

133 14 3 84 15 2 6 6 7 7

1 0 16 0 2 4 817 0 1 5 7 7

(b)

8 10 12 14 16 18

Pre

Pos

t

Number of inappropriate words for children injured before and after 15 years

Number of inappropriate words

(c) Both plots show the bimodal nature of the data in the over 15s and the bias to lower values inthe under 15s, however the dotplots are much harder to read. It is easier to read the data from thestemplots.

2.5 The stemplot (below) shows the data is bimodal, could indicate different types of driving altersthe car’s fuel consumption.

8 28 5 5 5 6 7 key9 3 3 4 8 | 2 = 8.29 6 6 7

10 1 1 2 310 5 5 6 6 7 8 911 0 1 311 5 6 6 6

2.6 We can see 2 distinct groups in the dotplot, with a large group around 80-85 and a smaller grouparound 87-89. There are 2 values, 73 and 76, that could be outliers. A possible explanation couldbe that for 2 of the incubation temperatures, there is no strong difference in SVL, with the 3rdgroup distinctly different. There could be other factors that affect SVL such as gender.

75 80 85 90

Snout vent lengths of 30 tuatara

Length (mm)

10

2.7

6 8 10 12 14

Maximum leaf lengths of 28 plants

Length of the largest leaf per plant (cm)

The dotplot shows clustering of leaf lengths around 6.5, 8 and 10cm. It is easier to recover thedata from the stemplot (below). Both plots show the shape of the data.

5 56 4 4 4 5 7 key7 2 8 8 6 | 4 = 6.48 1 3 3 4 7 89 6 7 9

10 1 3 711 4 412 713 1 314 2 5

2.8 The stemplot (below) shows plants from Carrick range are longer in general. Both are unimodaland the range for Carrick range is larger than that of Takahe Valley.

Carrick Range Takahe Valley

11 54 12 3 3 key

9 9 8 7 7 6 12 5 5 6 7 7 8 12 | 4 = 12.43 3 2 0 0 13 0 1 1 39 9 7 6 5 13 6

4 3 3 14

2.9 It is sensible to do a split-stem as most of the data is in the 10’s, there is not much shape to theplot if you don’t do a split stem as there are only 4 values in the stem. The stemplot is unimodaland quite symmetric, a representative value would be between 15 and 19.

0 8 91 0 4 4 4 key1 6 7 7 8 8 8 9 1 | 0 = 102 2 2 22 53 2 3

2.10 (a) The frequency table is:Class Boundaries Frequency

1.0-(1.5) 71.5-(2.0) 42.0-(2.5) 22.5-(3.0) 23.0-(3.5) 03.5-(4.0) 44.0-(4.5) 84.5-(5.0) 9

(b) The histogram is:

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Histogram of Soccer Player Fitness

Average number of sit−ups/12 seconds

Fre

quen

cy

1 2 3 4 5

02

46

810

02

46

810

0.5 1.5 2.5 3.5 4.5 5.5

(c) The data is bimodal, most soccer players could do a lot of situps or very few at all, with notmany in the middle.(d) The stemplot is more useful because you can see the shape of the data and no data are lost inthe stemplot, whereas the frequency table and histogram lose the actual data values.

1 1 1 2 3 3 4 41 5 6 6 8 key2 0 3 1 | 4 = 1.42 7 833 5 7 8 84 0 0 0 1 2 4 4 44 5 5 6 7 7 7 8 8 9

(e) Gender and Team are categorical. Average situps and Body weight are Quantitative continu-ous. Pulse rate is Quantitative Discrete.

2.11 The histogram (below) shows the data is unimodal with mode between 0 and 5%. The data isskewed to the left which means there is more data in the negatives, so monthly returns have moreoften been negative than positive over the five years of data.

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Histogram of Telecom Monthly Returns

monthly return (%)

Fre

quen

cy

−20 −10 0 10 20

05

1015

20

2.12

Histogram and Frequency Polygon of Intruder Words

words per 10−minute interval

Fre

quen

cy

6 8 10 12 14 16 18 20

02

46

810

13

2.13 (a) and (b)

Histogram and Frequency Polygon of Numbers of Pimelea concinna

number of plants (per 10 square m.)

Fre

quen

cy

0 10 20 30 40

02

46

8

−5 5 15 25 35

(c) Relative frequency. When comparing samples of different sizes, relative frequency must be used.(d) The data are discrete so a bar chart may be a better representation.

2.14 (a)

Class Boundaries Frequency0-(10) 1010-(20) 320-(30) 630-(40) 540-(50) 450-(60) 460-(70) 170-(80) 0

(b)

Histogram and Frequency Polygon of Numbers of Rostkovia magellanica

number of plants

Fre

quen

cy

0 20 40 60 80

02

46

810

10 30 50 70

(c) We would need to use relative frequency as the sample sizes are different.(d) Quantitative discrete.

2.15 (a) A histogram is more appropriate as the data is quantitative continuous. There are no gapsbetween class intervals.(b) The histogram (below) is skewed to the right with a mode in the size range 12.5 to 15cms.

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Note that around 40% of paua are undersize. There is also a smaller mode in the size range 20to 22.5cms, which suggests that there are 2 distinct groups in the paua population - this could bemale and female paua, or adult and juvenile.

Histogram of Paua Length

size of paua (cms)

Den

sity

5 10 15 20 25

0.00

0.02

0.04

0.06

0.08

0.10

0.12

7.5 12.5 17.5 22.5

2.16 (a) A histogram is appropriate as the data is continuous and there are no gaps in the class bound-aries.

Histogram of Availability of Family Support

availability of support

Fre

quen

cy

0 2 4 6 8 10 12

020

4060

80

(b) There are 2 modes, suggesting there are 2 distinct groups within the population of families inSouth Auckland when it comes to availability of family support.(c) 177 families got a score less than 4 and 222 got a score less than 5. The first mode occurs be-tween 3 and 4, so it is consistent with the statement. It is likely that the left hand side representsdysfunctional families.

(d) The stemplot is:

15

Dysfunctional Functional

5 08 1 key

2 6 | 0 = 6.09 3 1 39 8 6 4

7 5 58 4 6 0 8

7 5 88 1 7 9

3 9 0 2 4 85 10 2 5 9 9

(e) The data from the dysfunctional group has a large spread and the range covers nearly all pos-sible values, where the data from functional group is limited to scores of 6 and above. This mayindicate that there are factors other than availability of support that influence whether a family isdysfunctional.

2.17 (a)

Histogram of Chocolate Bar Weights

weights (grams)

Fre

quen

cy

49 50 51 52 53

02

46

810

12

(b) Visually, a stemplot is a rotated histogram, but the benefit of a stemplot is that you do notlose any information.

49 3 449 6 key50 1 2 3 4 4 50 | 1 = 50.150 6 7 7 7 8 8 8 8 8 951 0 1 1 2 2 3 3 3 451 5 5 5 6 6 7 7 7 8 8 852 252 9

(c) Most values lie within 50.5 and 52, so they usually get fairly close to the required of 50 grams.It is standard practice to aim slightly higher than specified weight, as there can be serious conse-quences for producing underweight products. 3 out of 40 (7.5%) have weight less than 50 grams,which may be too high depending on the company’s tolerance level.

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2.18 (a)

Organic Substance Schiz. Mood Neurotic Behav. Pers. Dis. Retard. Psych. Dev. Hyper.

Barplot of Diagnoses in New Zealand Mental Health

Diagnosis

% p

eopl

e

05

1015

2025

3035

(b) This can happen because a person may have more than one diagnosis recorded in the mentalhealth system in 2002, i.e. a person can appear in more than one bar, e.g. by having neurotic stressand anxiety, and being a substance abuser.

2.19 Showing more than one variable on a plot is difficult, unless scales are very similar in which casea combination of bars and lines might be possible. Here, we show the three temperature variablestogether, and the remaining three variables in separate barplots.

Average annual temperature with maximum and minimum

Tem

pera

ture

−5

0

5

10

15

20

25

30

35

Auc

klan

d

Gis

born

e

Tau

po

Nap

ier

Wel

lingt

on

Nel

son

Chr

istc

hurc

h

Dun

edin

Auc

klan

d

Gis

born

e

Tau

po

Nap

ier

Wel

lingt

on

Nel

son

Chr

istc

hurc

h

Dun

edin

Rainfall

Mea

n an

nual

rai

n da

ys

0

20

40

60

80

100

120

140

Auc

klan

d

Gis

born

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Tau

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lingt

on

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son

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istc

hurc

h

Dun

edin

Sunshine hours

Sun

shin

e ho

urs

per

annu

m

0

500

1000

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2.20 Attendance has loosely been from countries “close” to the venue. UK attendance has been muchhigher for the Southern European ICME; Chinese attendance much higher for the Japan ICMEthan the two European conferences; Brazilian attendance much higher for the Spain ICME. Anexception is the Australian attendance at the Japan ICME, which was low relative to its proximity.

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Spain Japan Denmark

MexicoSouth AfricaNew ZealandAustraliaBrazilUSAUKChinaItalyPoland

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2.22 The bar graphs (below) suggest that responses to the two questions are very similar but withquestion B getting more high responses than question A.

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1 2 3 4 5 6 7

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2.23 The bar graph shows the general decline of the Maori population in the Southern North Island,the exception being the Wairarapa, which was stable. Also clear is the large population in theWanganui region.

Port Nich. Waik/Porirua Otaki Manawatu Rangitikei Wanganui Wairarapa

1840s185018571874

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Chapter 3

Exercises

3.1 Using the data on brain injury in Exercise 2.4, find the mean and standard deviation of the numberof inappropriate words, for each of the groups injured before age 15 and injured after age 15.

3.2 Find the mean and standard deviation and variance of the leaf length data in Exercise 2.7.

3.3 Exports from New Zealand to Australia (in $million) over the four years 2000 to 2003 were: 5528,6083, 6326, 6050. Find the mean, median and standard deviation of these data.(Data from New Zealand in Profile, 2003 and 2004, Statistics New Zealand.)

3.4 Using your frequency table from Exercise 2.10:

(a) Write down the class midpoints and use them to estimate the sample mean and standarddeviation of the average number of sit-ups per 12 seconds to three decimal places.

(b) What characteristic of the data will cause the sample mean and median to differ? Which willbe larger and why?

3.5 Following from Exercise 2.5:

(a) Calculate the mean and standard deviation of the fuel consumption data.(b) Draw a boxplot of the data.(c) Why must the driver monitor km/L rather than km/$ in order to keep track of his fuel effi-

ciency?

3.6 Using the data from Exercise 2.22:

(a) Find the mean and median responses to Questions A and B. Which is it more appropriate touse and why?

(b) Find the range of responses in each case and also the interquartile ranges.(c) Summarise the information in (a) and (b) and compare the responses to questions A and B.

3.7 Using the histogram or frequency table from Exercise 2.14, calculate estimates for the true samplemean and sample standard deviation of the number of plants per site to two decimal places usingthe grouped data of the table.

3.8 Using the data from Exercise 2.13, calculate estimates of the true sample mean and sample standarddeviation of the number of plants per site to two decimal places using the grouped data of yourtable.

3.9 In an international test of secondary school mathematics achievement (P.I.S.A.) used in OECDcountries recently, New Zealand students had a very high mean score, 20% of New Zealand studentsscored in the 90th percentile and the range of scores in New Zealand students was among the veryhighest. Explain what this means in practical terms and why this is both good news and bad newsfor education officials in this country.

3.10 A random sample of oil spills in the Southern hemisphere (in tonnes of oil × 100,000) is as follows:1.53, 0.48, 0.42, 0.34, 1.76, 0.43, 0.33, 0.19, 0.5, 1.1, 0.1, 0.57, 0.36, 0.20, 0.23, 0.64, 0.20, 0.18, 0.6,1.0. Find the mean and variance of oil spill sizes from this sample.

3.11 Using the data and histogram from Exercise 2.12, write down the class midpoints and then estimatethe sample mean and sample standard deviation of the number of intruder words. Also estimatethe range of the data. Explain why these are estimates only. Summarise the main features of thisdata.

3.12 (a) Find the median, quartiles and the 85th percentile of the data from the Takahe Valley inExercise 2.8.

(b) Are any of the data from the Takahe Valley to be considered outliers? Give reasons for youranswer.

21

(c) Draw horizontal boxplots for the data from both the Takahe Valley and the Carrick Range andwrite a sentence comparing the two boxplots. What does this suggest about the two locations?

(d) Draw a modified boxplot of the Takahe Valley data.

3.13 A survey of the cost of energy generation in New Zealand power stations conducted in 1985-86indicated the following total operating costs (cents/kW):

Station Cost Station Cost Station Cost

Ohau A 0.05 Aratiatia 0.43 Cobb 1.26Ohau B 0.07 Matahina 0.46 Tuai 1.29Ohau C 0.11 Tokaanu 0.56 Monowai 1.92Tekapo B 0.11 Ohakuri 0.58 Lake Coleridge 2.11Roxburgh 0.15 Arapuni 0.58 Mangahao 2.26Manapouri 0.16 Waitaki 0.58 Huntly 3.43Piripaua 0.16 Atiamuri 0.59 New Plymouth 3.48Rangipo 0.18 Waipapa 0.60 Stratford 3.85Benmore 0.20 Highbank 0.77 Otahuhu 7.45Aviemore 0.21 Karapiro 0.79 Meremere 15.52Maraetai 0.35 Wairakei 0.91 Whirinaki 20.61Whakamaru 0.39 Arnold 1.23 Marsden A 55.99Kaitawa 0.42

(a) Find the mean and standard deviation of the total operating costs per kW. Why are these notsensible summary statistics for these data?

(b) Draw a modified boxplot of the data, and comment on any features.

(Data thanks to Richard Martin.)

3.14 The following prices (in $000) are of 27 central city apartments which were among those sold inWellington in early 2004.

277.5 290 320 330 340 350 359 359 370 385400 400 460 475 493 500 550 570 590 600600 628 635 655 667 1260 1700

Find the sample mean and sample median of these data. Which is more appropriate?

(Data thanks to Ron Beccard.)

3.15 In Exercise 2.16:

(a) What was the approximate range of availability of family support scores observed in thissample? Explain why it is an estimate only.

(b) Calculate estimates of the sample mean and standard deviation of the data to two decimalplaces.

3.16 A botanist follows a transect along a hillside. For each 12 totara trees she passes, she records howmany out of the 12 are still in the juvenile stage. Later, this data is collected into a frequency tableas follows:

Number of juveniles (out of 12) 0 1 2 3 4 5 6 7 8 9 10 11 12

Frequency (number of samples) 7 15 20 11 6 0 1 1 1 1 0 0 0

How many totara trees were looked at altogether? Find the mean number of juveniles in groupsof 12 totara trees and the standard deviation. What would be an appropriate graph in which todisplay this data?

3.17 Calculate the sample mean and standard deviation of the monthly returns for Telecom given inExercise 2.11.

22

3.18 The following data was collected by Cielle Stephens on the New Zealand spotted skink. Thirty-twotraps were placed in each of three habitats: pasture, replanted forest and tussock on StephensIsland. The data are the counts of skinks per trap totalled over a ten-day period in each habitat.

Pasture 4 3 0 2 2 1 4 1 2 5 0 1 5 6 5 611 3 1 1 4 8 5 14 6 8 10 7 4 8 13 6

Replanted 15 24 31 8 4 18 14 33 11 16 20 1 17 12 27 26forest 18 6 12 16 11 8 13 12 11 8 10 17 29 3 12 5

Tussock 14 23 15 14 5 16 10 16 14 10 7 10 8 12 19 177 12 29 10 11 11 10 10 6 13 7 10 8 12 6 12

Draw side-by-side boxplots of this data and summarise what they are saying in terms of skinkdensity with respect to these three habitats.

3.19 The closing prices of the 50 stocks in the NZSX 50 Index on Friday 2 July, 2004 were (in $):

6.82 0.41 6.88 20.15 0.84 3.21 0.67 2.06 4.90 5.86 0.85 0.94 4.214.60 4.53 13.08 2.44 1.69 1.96 4.99 2.95 3.15 4.68 1.06 7.46 18.555.94 2.89 5.30 0.94 4.50 7.05 5.20 2.11 0.82 1.25 2.80 4.60 4.784.75 1.30 5.18 4.14 5.85 1.90 1.78 5.48 4.65 1.71 0.95

(a) Draw a stemplot of these data and comment on any features.(b) Draw a modified boxplot of the closing prices.(c) Calculate the sample mean and standard deviation of these prices. Repeat with the outliers

removed, and comment on your estimates.

3.20 For 18 countries in Central and South America the following birthrates (births per 1000 population)were reported in 1996:

38 34 39 38 23 25 28 19 2739 29 25 37 31 34 24 30 26

(a) Find the median birthrate for this group of countries in 1996 and the 35th percentile.(b) By finding the upper and lower quartiles and maximum and minimum, draw up a boxplot of

this data.(c) Should any of the data be considered outliers? Why?(d) Find the mean and standard deviation of this data. By comparing the mean and median what

characteristic of the data can you deduce?

3.21 The labour force participation rate (%) in over-15-year-olds in the main urban areas of the NorthIsland in 2001 were: 60.6, 68.3, 69.4, 67.3, 66.2, 68.5, 65.9, 64.4, 61.1, 66.8, 62.3, 63.8, 64.8, 61.9,59.2, 66.1, 56.2, 66.8, 69.2, 69.3, 77.4.

(a) Find the mean, median and standard deviation of this data.(b) Are there any outliers in this data?(c) Draw a modified boxplot of the data.

(Data from 2001 Census Regional Summary, Statistics New Zealand.)

3.22 The survival times (in months) for two groups of patients with a terminal illness, one group with23 patients who are given the standard treatment, and another group of 19 patients who are givenan experimental treatment are shown below:

Standard treatment Experimental treatment

9 13 13 16 17 17 19 27 28 35 38 3921 24 25 25 31 40 44 44 46 47 54 5743 50 50 51 54 56 58 58 59 63 65 6556 60 62 62 64 68

23

(a) Find the lower and upper quartiles, the interquartile range and the range for each group.(b) Are there any outliers in either group? Justify your answer.(c) Draw side-by-side boxplots of the data for the two treatments. Summarise the information in

the boxplots.

3.23 The number of days that two different batteries for calculators last are listed below:

Battery A 614 710 648 768 868 724 548Battery B 590 672 844 658 778 748 946 652 772

Construct side-by-side boxplots for each battery type and summarise the information in the plots.

3.24 The annual rainfall (in mm) over 12 South Island regions in the year 2001 was:

47 56 116 178 19 75 160 31 12 164 43 74

Find the mean, range, standard deviation and variance of this data.

3.25 Below are the lengths of time until reoffending of a sample of repeat offenders who are all malesof similar age and with a similar history of offending. The time until reoffending is recorded inmonths according to the type of sentence they had last received.

Jail with job training 14 3 16 10 11 12Jail only 18 15 21 14 16

(a) Find the mean time until reoffending for each category of offender and their associated standarddeviations.

(b) Find the median times until reoffending of each category of offender and the upper and lowerquartiles.

(c) Find the interquartile range in each case and use it to establish whether there are any outliersin the data.

(d) Draw side-by-side modified boxplots of the data and use them to summarise the informationin the data.

3.26 In a culture of a certain bacterium, the following counts in each of 100 samples of the same areawere recorded:

Number in area 0 1 2 3 4 5 6Frequency 8 17 28 23 16 5 3

Find the mean number of bacteria per sample and the standard deviation.

3.27 The following data represent the average total annual incomes of a random sample of superannuatedcouples.

Total income ($000) Number of couples

5 - 9.999 210 - 14.999 2715 - 19.999 5920 - 24.999 6025 - 29.999 2830 - 39.999 1840 - 49.999 550 - 99.999 1

(a) Find the approximate mean annual income of such couples and the associated standard devi-ation.

(b) Find an approximate median annual income for such couples.

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3.28 The assets of 42 companies in 2003 (in $million) are shown in the following table:

Assets 0-(2) 2-(4) 4-(6) 6-(8) 8-(10) 10-(12) 12-(14) 14-(16) Total

Frequency 3 5 5 7 10 8 3 1 42

Calculate estimates for the sample mean and standard deviation of the assets in $million using thisgrouped data.

3.29 Calculate the mean, standard deviation and 10th percentile of the chocolate bar weights in Exer-cise 2.17. Approximate the mean and standard deviation from the histogram and compare theseto the actual values.

Solutions

3.1 Injured before 15: x = 10.209, s = 2.867. Injured after 15: x = 13.85, s = 3.230.

3.2 x = 9.364, s = 2.539 and s2 = 6.445.

3.3 x = 5996.75, median= 6066.5 and s = 335.860.

3.4 (a) The midpoints are 1.25, 1.75, 2.25, 2.75, 3.25, 3.75,4.25, 4.75. x = 3.264, and s = 1.401.(b) When the data is skewed, the sample mean and median will differ. In this case, most of thevalues are large, so the fewer low values will bring the mean downwards, without affecting themedian (see answer to 2.10).

3.5 (a) x = 10.097, and s = 1.055.(b) min = 8.2, max = 11.6, LQ = 9.3, median = 10.25 and UQ = 10.9

8.5 9.0 9.5 10.0 10.5 11.0 11.5

Boxplot of Fuel Consumption

fuel consumption (km/L)

(c) The price the driver is paying for fuel can vary over time which would make it appear thatefficiency is changing even when the driver is still getting the same kms per litre.

3.6 (a) Question A: x = 3.1 and median= 3. Question B: x = 3.7 and median= 4. The median andmean are similar in both cases, expect symmetric distribution, so mean is more useful since it usesall of the data.(b) Question A: range = 7− 1 = 6, IQR = UQ−LQ = 4− 2 = 2. Question B: range = 7− 1 = 6,IQR = UQ − LQ = 5 − 2.5 = 2.5(c) Both questions yielded fairly symmetric distributions, with Question A having a lower meanthan Question B. The ranges for both questions are the same and IQRs are similar. People in thetest gave higher values in the Likert scale for seeing the event, compared with hearing the event.

3.7 x = 26.82 and s = 18.95 (2dp).

3.8 x = 12.83 and s = 9.09 (2dp).

3.9 Since the range is one of the highest, even though many did very well, there must be a lot ofstudents who also did very poorly. On the other hand one-fifth of students were in the top 10%internationally. It will make it difficult to target education to a particular level.

3.10 x = 0.558 and s2 = 0.206.

3.11 The midpoints are 7.5, 8.5, 9.5, 10.5 11.5, 12.5, 13.5, 14.5, 15.5, 16.5, 17.5. The approximate meanis x ≈ 12.683 and s ≈ 3.634. These are estimates because we are using the midpoint of each class

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instead of the actual data. Range ≈ 18−7 = 11, and this is an overestimate because we don’t knowthe actual max and min. The average number of intruder words is approximately 12.68 but thisis misleading as the data is bimodal and there are two peaks, one around 9.5 and another around17.0. The range of intruder words is from around 7 up to 18 so there is quite a spread but 40% ofthe data is in the group 15-(18).

3.12 (a) n = 14, np = 14 × 0.5 = 7, so median between 7th and 8th values: median= 12.7. p = 0.25:np = 14 × 0.25 = 3.5, so 4th value: LQ = 12.5. p = 0.75: np = 14 × 0.75 = 10.5, so 11th value:UQ = 13.1. p = 0.85: np = 14 × 0.85 = 11.9, so 12th value: 85th percentile = 13.1.(b) IQR = UQ − LQ = 13.10 − 12.5 = 0.6. UQ + 1.5IQR = 14.00, LQ − 1.5IQR = 11.6. 11.5 isthe only value outside these values, so it is an outlier.(c) Plants from Carrick Range are generally longer, both ranges are similar. Takahe Valley is moreskewed.

Car

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Side by Side Boxplots for Lengths of Rostkovia magellanica

plant lengths (cm)

(d) Note that the left hand whisker ends at 12.3 - the smallest value that is not an outlier.

11.5 12.0 12.5 13.0 13.5

Modified Boxplot for Lengths of Rostkovia magellanica in Takahe Valley

plant lengths (cm)

3.13 (a) x = 3.508 and s = 9.807. These are not sensible summary statistics because of the shape ofthe distribution. It is asymmetric, and the few stations with high costs have a large influence overthese statistics, which don’t represent the bulk of the data. They have the effect of dragging themean well above the ‘middle’ value.(b) LQ = 0.21, UQ = 1.92, IQR = 1.71, LQ − 1.5IQR = −2.355 and UQ + 1.5IQR = 4.485. Theboxplot shows four outliers at the higher costs, and also shows that the data is very asymmetric.

0 10 20 30 40 50

Modified Boxplot for Energy Generation Costs

total operating costs (cents/kW)

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3.14 x = 539.4 (1dp) and median = 475. The median is more appropriate due to the outlying values.

3.15 (a) Range ≈ 11 − 0 = 11, which is approximate because we do not know the actual data and canonly use the minimum of the first and maximum of the last groups with data in them.(b) x ≈ 5.52 and s ≈ 2.73.

3.16 The sample size is∑

fi = n = 63. The total number of trees is 12× 63 = 756 trees. x = 2.254 ands = 1.796. A barplot would be appropriate since the data are discrete.

3.17 x = −0.727 and s = 6.352 (3dp).

3.18 Pasture: min = 0, LQ = 2, median = 4.5, UQ = 6.5 and max = 14.Replanted: min = 1, LQ = 9, median = 12.5, UQ = 18 and max = 33.Tussock: min = 5, LQ = 9, median = 11, UQ = 14 and max = 29.

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Side by Side Boxplots for Skink Density

number of skinks per trap

The boxplots show that the skink density in tussock and replanted forest are similar, the mediansare approximately the same with the range in tussock slightly smaller. The skink density in pastureis lower than the other 2 groups, so must be the least favoured habitat for skinks.

3.19 (a) Note how difficult it is to read a stemplot with this data. A boxplot or histogram is much moreinformative.

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04 1 42 1 key05 : 08 | 2 = 0.8206 7 45 0 307 46 0 0 5 808 2 4 5 47 5 809 4 4 5 4810 6 49 0 911 5012 5 51 813 0 52 0

: 53 016 9 54 817 1 8 :18 58 5 619 0 6 59 420 6 :21 1 68 2 8

: 6924 4 70

: :28 0 9 74 629 5 :30 130 831 5 :32 1 185 5

: :41 4 201 5

(b) min = 0.41, max = 20.15, median = 4.175, LQ = 1.71, UQ = 5.2, IQR = 3.49, LQ−1.5IQR <0 and UQ + 1.5IQR = 10.435. Thus 13.08, 18.55 and 20.15 are outliers.

0 5 10 15 20

Modified Boxplot of NZSX prices

Stock closing price

(c) x = 4.296 and s = 3.926. Removing the outliers: xadj = 3.469 and sadj = 2.039. By removingthe outliers, both the mean and standard deviation decrease markedly.

3.20 (a) median = 29.5, 35th percentile = 27 (np = 18 × 0.35 = 6.3, so 7th value).(b) LQ = 25, UQ = 37 , min = 19 and max = 39.

20 25 30 35

Boxplot of Birthrates

birthrates (births per 1000 pop.)

(c) IQR = 37 − 25 = 12, LQ − 1.5IQR = 7 and UQ + 1.5IQR = 55. There are no data outsidethis range, so there are no outliers.(d) x = 30.333 and s = 6.212. The mean and median are slightly different, so you would expectthe distribution to be skewed.

3.21 (a) x = 65.5, s = 4.5098, median is the 11th piece of data when ordered = 66.1.(b) 0.25n = 5.25 so the LQ is the 6th score = 62.3, and UQ is the sixth from the top = 68.3.IQR = 68.3 − 62.3 = 6, UQ + 1.5IQR = 77.3 and LQ − 1.5IQR = 53.3. Only one observation isoutside these limits, 77.4, and this is an outlier.

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(c)

60 65 70 75

Modified Boxplot of Labour Force Participation

labour force participation (%)

3.22 (a) Standard treatment: LQ = 17, UQ = 56, IQR = 56 − 17 = 39 and range = 64 − 9 = 55.Experimental treatment: LQ = 38, UQ = 59, IQR = 59 − 38 = 21 and range = 68 − 19 = 49.(b) Standard treatment: LQ−1.5IQR = −41.5 and UQ+1.5IQR = 114.5, there are no outliers forthe standard treatment group. Experimental treatment: LQ− 1.5IQR = 6.5 and UQ + 1.5IQR =90.5, so there are no outliers in this group.(c) The survival times are more spread out for standard treatment, and more people survive longerwith the experimental treatment. The experimental treatment has a much narrower IQR. Bothdistributions are skewed.

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Side by Side Boxplots for Different Treatments

survival times (months)

3.23 Battery A: min = 548, LQ = 614, median = 710, UQ = 768 and max = 868.Battery B: min = 590, LQ = 658, median = 748, UQ = 778 and max = 946.

AB

600 700 800 900

Side by Side Boxplots for Battery Lifetimes

battery lifetime (days)

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Battery B appears to last longer than Battery A due to a higher median. A has a wider interquartilerange than B, i.e. more variable lifetimes.

3.24 x = 81.25, range = 178 − 12 = 166, s = 58.913 and variance, s2 = 3470.75.

3.25 (a) For those with job training x = 11 and s = 4.472, while for those with jail only x = 16.8 ands = 2.775.(b) In first group, ordered data is: 3, 10, 11, 12, 14, 16. Median is 11.5, LQ = 10 and UQ = 14.In second group ordered data is: 14, 15, 16, 18, 21 with median 16, LQ = 15 and UQ = 18.(c) IQR for job training group = 14 − 10 = 4 and 1.5 × IQR = 6. UQ + 1.5IQR = 14 + 6 = 20.LQ − 1.5IQR = 10 − 6 = 4. There is one data value outside of 4 and 20, so 3 is an outlier in thisgroup. For the jail only group IQR = 18−15 = 3 and 1.5×IQR = 4.5. UQ+4.5 = 18+4.5 = 22.5while LQ − 4.5 = 15 − 4.5 = 10.5. There are no outliers in this group.(d) Those that had training appear to reoffend sooner than those in jail only.

with

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Side by Side Boxplots for Reoffending Times

reoffending time (months)

3.26 x = 2.49 and s = 1.439.

3.27 (a) Class midpoints are 7.5, 12.5, 17.5, ..., 45, 75. (Note: classes are not all the same width.) Usingthese, x ≈ $22175 and s ≈ $8207.1.(b) n = 200 couples, so median is the average of the 100th and 101st when listed in order. Thesum of the frequencies in the first 3 classes is 2+27+59 = 88 which is less than 100 so the medianmust be in the next class where we need 12 more couples to get 100 couples. So the 100th score isapproximately 12

60of the distance from 20 to 25 thousand dollars. This class has a range of $5000

and multiplying $5000 by 12/60 we get $1000 so we need to add $1000 to the lowest boundary of$20,000 to get the 100th score at roughly $21,000. The 101st will be $5000/60 = $83.33 more, andthe average of these is $21041.67.

3.28 x ≈ $7.714m and s ≈ $3.584m.

3.29 x = 51.038 and s = 0.735, for the 10th percentile np = 40 × 0.1 = 4, so will be half way between4th and 5th value of the sorted data, which is 50.15. Using grouped data from the histogram, with8 intervals, x ≈ 51.0625 and s ≈ 0.757. These values don’t vary much from the actual values, thelength of class intervals (0.5 grams) are small and so the difference between midpoints and dataare small. (NB: these values will vary slightly depending on the choice of class intervals.)

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Chapter 4

Exercises

In these first exercises, we practice drawing scatterplots, and examining the shape of the data.

4.1 As part of a battery of psychological tests given to eight subjects, two tests of spatial reasoningwere given, one a test of ability to solve geometric problems which yielded a score, x and the secondin which subjects were tested on their ability to judge distances, which gave a score, y. The resultswere:

Subject A B C D E F G H

Test I score, x 115 101 89 112 105 90 91 88Test II score, y 20 25 21 22 23 22 24 22

Draw a scatterplot of these scores. What does your plot suggest about the data?

4.2 The health of a lake in the central North Island is being monitored at eight sites by sampling thedissolved oxygen level and the frequency of a trout species with the following results:

Site A B C D E F G HMean dissolved oxygen level, x 27 12 25 26 16 15 23 16Number of trout in catchment, y 42 17 35 43 27 20 39 27

Draw a scatterplot of this data and use it to comment on any relationship that seems to existbetween oxygen level and numbers of fish at a site.

4.3 In a study of reaction times (ms) to two particular images shown on a computer screen (oneconventionally pleasant and one unpleasant) with ten subjects the following data was collected:

Pleasant image, x 6.5 4.5 2.0 5.0 4.5 13.0 7.4 4.5 15.9 4.0Unpleasant image, y 5.7 8.6 1.9 3.7 2.2 7.8 3.4 2.2 4.0 3.4

Draw a scatterplot of these data. What does the scatterplot suggest about the relationship (if any)of the reaction times to a pleasant image compared to that of an unpleasant image?

4.4 As part of a study of the effect of group pressure on 10 subjects two tests were given. One wasa test of compliance to orders given by an authority figure (x) and the other was a score from asimulated accident in which the subject was scored on willingness to be involved (y):

x: 87 105 94 123 120 124 89 132 113 124y: 40 44 37 63 86 90 38 93 52 79

Sketch a scatterplot of this data. What appears to be the relationship between the two scores?Does it seem that there may be a linear relationship between these two test scores?

4.5 The following data is on the purchase of Maori land in Wellington province from 1839 to 1876:

Region Wairarapa Wellington Wanganui Rangitikei Manawatu

Amount acquired (103 acres) 3428 260 283 287 636Price/acre (pence) 5.2 13.6 12.0 12.6 13.9

Draw a scatterplot of this data. Does there appear to be a relationship between these two variables?If so, what is it?

(Data thanks to Brad Patterson, from McConchie, Winchester and Willis, Dynamic Wellington(2000) Institute of Geography, Victoria University.)

31

In this next set of problems, we continue to draw scatterplots, and we also calculate correlation coeffi-cients, and estimate regression lines.

4.6 Nine young adolescent women with repeated truancy from school were given several psychologicaltests. In particular, their anxiety level, x, was recorded, and compared to their scores on a testmeasuring self-esteem, y. The following data was obtained:

Subject A B C D E F G H I

Anxiety level 2 5 6 10 10 17 30 10 8Self-esteem 5 20 22 30 26 35 40 30 27

(a) Plot a scatterplot of these data.(b) Which of Pearson’s or Spearman’s correlation coefficients seems more appropriate here? Give

reasons for your answer.(c) Calculate Pearson’s linear correlation coefficient to 3 decimal places.(d) Rank the scores for each test and calculate Spearman’s correlation coefficient for this data.(e) What do you conclude about the relationship between anxiety and self-esteem in this group of

young women?

4.7 List the five correlation coefficients in order, from that which indicates the weakest linear relation-ship to that which indicates the strongest linear relationship: 0.6,−0.8, 0.5,−0.4, 0.0.

4.8 (a) Draw a sketch of a scatterplot of a population that has a correlation of approximately zerobetween two characteristics X and Y .

(b) On the same sketch identify a sample of that population that will have a correlation betweenX and Y of approximately 1.

4.9 The following statements all contain errors. Identify and explain the error(s) in each case.

(a) There is a moderately weak correlation of r = 0.3 between men’s and women’s wages in NewZealand.

(b) The correlation between Nick Willis’ times for the 1500m and 3000m is a strong r = 1.21.(c) The correlation between length of prison sentence and ethnicity in Australia is r = 0.85.(d) A weak negative correlation of r = 0.18 exists between height and occupation.(e) There is an extremely weak correlation of r = −0.94 between the duration of dementia with

the degree of cognitive functioning.

4.10 In a test of the reliability of memory a random sample of ten subjects aged between 18 to 25 wereseated in a room in which they believed they were about to be involved in a stimulus-reactionexperiment. An eleventh person was in the room typing at a computer. Another “student” cameinto the room and angrily demanded that the eleventh student get off the computer using threatsof violence and eventually left after appearing to hit the other who yelled loudly and left the roomdoubled up. The whole episode was videotaped. The ten subjects were then asked about theincident they had witnessed. Their recollection was scored for accuracy and their scores on anassessment of spatial ability were also recorded. The researcher was interested in whether accuracyof recollection in a three-dimensional setting is related to spatial ability. The data were as follows:

Spatial ability 8.9 7.9 7.6 9.2 9.1 6.3 8.6 6.4 7.6 6.5Accuracy of recollection 60 56 53 71 68 50 59 47 53 42

(a) Draw this data on a scatterplot and calculate Pearson’s correlation coefficient for the data.(b) Comment on the correlation coefficient obtained in (a). What does it suggest?(c) On inspection it seems likely that the accuracy of the recollection scores are rather doubtful,

but we can at least be sure that a larger score is genuinely larger than a smaller one, i.e. 59 isgreater than a score of 50, etc. Calculate Spearman’s correlation coefficient for the data. Whyis it appropriate in this situation?

32

4.11 The level of dopamine in children with petit mal epilepsy is thought to be related to the number ofepisodes of petit mal in a day. Eight such children were closely monitored over a two-week periodand for these children the following data was collected on their frequency of episodes where 1 =almost none, 2 = occasional, 3 = regular, 4 = very frequent:

Dopamine level, x 6.5 8.3 7.8 9.7 8.1 7.5 7.0 6.1Episodes of petit mal, y 1 4 3 4 3 2 2 1

Compute Spearman’s correlation for this data. Why is it the appropriate correlation coefficient forthis data?

4.12 In the same experiment discussed in Exercise 2.22, two further questions given to the studentsshown the sequence of photographs were: C: “As I remember the event, I can feel the emotionsthat I felt then”, and D: “I believe the event in my memory really occurred in the way I rememberit and that I have not imagined or fabricated anything that did not occur.” The scores of eightstudents on these questions for Question C where 1 = not at all and 7 = as clearly as if it werehappening right now, and for Question D where 1 = totally imaginary and 7= 100% real, are asfollows:

Question C 2 5 3 5 7 6 5 3Question D 6 5 6 3 7 5 6 7

(a) Calculate a Spearman’s correlation coefficient for the responses to the two questions.(b) What does your answer imply?(c) Why is Spearman’s an appropriate correlation coefficient to use in this situation?

4.13 Individuals from a business training programme are monitored in their first year in business, andtheir achievement in that first year quantified using a range of tests. The training programme wasbased on a mentoring system, with young trainees paired in every case with older, more experiencedpeople who were training for a career change. The data for the individuals are:

Team 1 2 3 4 5 6 7 8

Young Trainee 30 44 21 17 25 22 16 28Mentor 23 37 7 18 10 15 13 29

(a) Draw a scatterplot of the data, using the mentor’s scores to predict the young trainee’s scores.(b) Calculate Pearson’s correlation coefficient for the data, and interpret your figure.(c) Now calculate Spearman’s correlation coefficient for the data.(d) Which correlation do you think is most appropriate in this situation?

4.14 The metabolic rate of animals is often measured by oxygen consumption in a closed chamber.Resting metabolic rate was measured in the Marlborough Green Gecko (Naultinus manukanus)over five trials on successive days, to find if the animals needed conditioning to accustom them tothe chamber before resting metabolic rate could be reliably measured. The following partial dataset gives the metabolic rate readings for the first and second trial at which each individual settledenough for a reading to be taken. The measurement is volume (in mL) of oxygen per hour per unitbody mass (in g). Vol1 represents the first trial, Vol2 the second.

Gecko 1 2 3 4 5 6 7 8 9 10

Vol1 0.150 0.061 0.126 0.098 0.109 0.076 0.130 0.062 0.075 0.099Vol2 0.132 0.058 0.105 0.078 0.080 0.085 0.103 0.068 0.074 0.080

Gecko 11 12 13 14 15 16 17 18 19 20

Vol1 0.073 0.147 0.150 0.063 0.086 0.079 0.192 0.084 0.102 0.090Vol2 0.076 0.067 0.149 0.060 0.080 0.078 0.140 0.064 0.107 0.097

(a) Draw a scatterplot of this data.

33

(b) Find Pearson’s correlation coefficient for the relationship between the metabolic rate of thegeckos in the first trial to that in the second.

(Thanks to Kelly Hare and Shirley Pledger for this data.)

4.15 In his work with sufferers of Parkinson’s Disease, Jared Smith obtained the following data fromnine sufferers on the duration of the disease (in years) and its severity (on the Hoehn and Yahrscale of disability):

Duration 6 1 3.5 1.5 2.5 18 11 8 3Severity 2 2 2 2 3 4 3 3 2

Calculate Spearman’s correlation for this data. What does the result imply?

(Thanks to Jared Smith for this data.)

4.16 Nine randomly selected subjects of the same sex were questioned on the degree of anxiety they feltabout their romantic attachments, e.g. by responding to items such as “I rarely worry about mypartner leaving me,” and also on their avoidance score which is a score relating to people’s comfortwith closeness in relationships, using items such as “I prefer not to show a partner how I feel deepdown.” The data is below:

Anxiety score, x 2.17 0.89 3.11 3.11 2.28 3.00 0.94 2.89 3.94Avoidance score, y 0.89 1.11 2.5 2.72 3.17 2.61 2.39 2.44 1.78

Calculate both Pearson’s correlation coefficient and Spearman’s correlation coefficient for this data.Draw a scatterplot of the data and use it to decide which of these two correlation coefficients seemsmore appropriate. Give a reason for your decision.

(Data thanks to Chris Sibley.)

4.17 The following is data from 12 randomly selected Pakeha subjects of the same sex giving theirsocial dominance orientation score (SDO) and their support for biculturalism in principle (SBiP).SDO measures belief in group superiority versus egalitarianism. Both scores were arrived at fromresponses to a number of items on a Likert scale. Scores closest to 5 are high, to 0 are low.

SDO score, x 1.07 2.43 2.21 2.07 3.14 2.14 2.14 2.79 2.57 2.00 2.50 2.43SBiP, y 3.6 3.4 3.8 3.4 3.8 3.6 2.0 4.0 3.0 2.8 3.4 3.2

Calculate Pearson’s correlation coefficient for this data. Comment on its value.

(Data thanks to Chris Sibley.)

4.18 The following data are the percentage returns on the top ten New Zealand stocks (NZSX10), andalso for the Australian Stock Exchange (ASX) for 1990-2004.

Year 1990 1991 1992 1993 1994 1995 1996 1997

NZSX10 −0.06 −0.25 0.02 0.04 0.26 0.03 0.02 0.14ASX 0.00 0.00 0.08 0.05 0.16 −0.01 0.13 0.18

Year 1998 1999 2000 2001 2002 2003 2004

NZSX10 −0.22 0.00 −0.02 −0.07 0.07 0.03 0.10ASX −0.04 0.15 0.04 0.07 −0.05 −0.05 0.15

(a) Draw a scatterplot of these data.(b) Calculate Pearson’s correlation coefficient for this data. Comment on its value.

4.19 Using the data for the ranking of the top five container ports in 2001 and 2002 from Exercise 2.21,evaluate Spearman’s correlation coefficient for the data for these two years.

The following set of exercises are for regression, and include revision of scatterplots and correlation.

34

4.20 The levels of heavy metals in seas are an important measure of the quality of the water and itsability to sustain fish life. Recent studies of the Baltic Sea have measured the amounts of strontiumand lead (as a % of the total weight of the sample). Natural levels of these metals in the Balticare not known but recent sampling yielded the following data:

Weight of strontium, x 4.0 5.5 4.5 4.25 6.0 5.75 4.5 4.0Weight of lead, y 14 17 19 14.25 18 22 16 16

(a) Plot a scattergram of these data.(b) For this data calculate the slope, b and the intercept, a of the fitted regression line. Write

down the equation of the line and plot it on your scatterplot.(c) Use your regression line to predict the weight of lead, y when the weights of strontium are

x = 4.0 and x = 6.0.(d) Why is it not sensible to predict the level of lead when the weight of strontium is 7.0?

(Data thanks to Ross Renner.)

4.21 Residential house sales in the Wellington suburb of Karori in the first half of 2004 included thesample below. The sale price for each home was recorded (in $000), as well as the number ofbedrooms.

Price Bedrooms Price Bedrooms Price Bedrooms Price Bedrooms

128 1 199 3 214 3 365 4144 2 201 3 215 3 315 5190 2 203 3 235 4 335 5195 2 205 3 290 4 280 6203 2 208 3 310 4 289 6182 3 210 3 340 4 505 7

(a) Plot these data using a scatterplot.(b) Estimate Pearson’s correlation for these data.(c) Estimate the regression line for the data, using the number of bedrooms as the predictor

variable.(d) Predict the sales prices of homes in Karori with 1, 2, 3, 4, 5, 6, and 7 bedrooms.

(Data thanks to Ron Beccard.)

4.22 Using the data in Exercise 4.10:

(a) Estimate the intercept and slope of the fitted regression line to this data. Plot this fitted line.(b) Estimate the proportion of the variability in the recollection scores that is explained by the

regression line?(c) Use the regression line to predict the accuracy of recollection score for a subject with a spatial

ability score of 6.8.

4.23 Using the data in Exercise 4.13:

(a) Estimate the regression line for these data.(b) Use the regression line to predict the score for a young trainee whose mentor got a score of 25?(c) Another mentor gets a score of 45. Why does the regression line not provide a reliable score

for their young trainee in this case?

4.24 Using the data in Exercise 4.14:

(a) Estimate the intercept and slope of the fitted regression line for the data.(b) Use the regression line to predict the metabolic rate for a gecko in trial 2 who had a metabolic

rate of 0.070 in trial 1.

4.25 An online store has constant sales of its major product throughout the year. The manager haskept track of her monthly advertising expenditure and the profits (both in $000) for this productover the last year. The results are:

35

Advertising 0.5 0.7 0.65 0.8 0.8 0.7 0.5 0.6 0.8 0.45 0.5 0.6Profit 21.7 25.9 24.1 28.3 28.9 27.0 22.6 22.7 28.3 21.3 21.8 24.1

(a) Draw a scatterplot of the data, using advertising as the independent variable.(b) Estimate the regression line for this data.

4.26 Using the data from Exercise 4.17 find the equation of the regression line using social dominancescore (SDO) as the explanatory variable. Use your regression line to predict support for bicultur-alism in principle scores for individuals with SDO scores of 2.5 and 3.0 respectively.

4.27 Below are data on the median income per region, x (Income) and the percentage of white collarworkers per region y (%) in New Zealand.

Region Income % Region Income %

Northland 15200 30.9 Wellington 22400 46.4Auckland 21100 43.1 Tasman 16100 27.1Waikato 18100 32.2 Nelson 17100 36.0Bay of Plenty 16800 33.2 Marlborough 17000 28.5Gisborne 15400 30.0 West Coast 14500 28.1Hawke’s Bay 16700 29.7 Canterbury 17600 36.1Taranaki 17300 29.7 Otago 15700 33.4Manawatu-Wairarapa 16300 32.1 Southland 17300 25.4

(a) Draw a scatterplot of this data and calculate Pearson’s correlation coefficient and r2.(b) What is the meaning of the r2 value?(c) Do any points on your scatterplot stand out as potential outliers? Which ones?(d) Calculate the least squares regression line for this data and plot it on your scatterplot.(e) Use your regression line to predict the percentage of white collar workers in a region with a

median annual wage of $17,500.

(Problem from Richard Willis. Data from 2001 Census: Regional summary, Statistics New Zealand.)

4.28 Using data from Exercise 4.3:

(a) Find the correlation coefficient and the equation of the regression line for the study of reactiontimes.

(b) What proportion of the variability in reaction times to the unpleasant images can be explainedby the reaction times to the pleasant images?

(c) Predict the reaction times to unpleasant images for subjects whose reaction times to the pleas-ant image were 5, 10 and 15 respectively.

4.29 In Exercise 4.4 you were asked to draw a scatterplot of compliance to orders given by an authorityfigure (x) versus willingness to help in a simulated accident (y). Compute the intercept and slopeof the least squares regression line for this data and draw on the scatterplot. Does the line appearto be a good fit to the data?

4.30 Exercise 4.2 provides data on the health of a North Island lake. Find the regression line for thisdata and predict the number of trout in the catchment for mean dissolved oxygen levels of 20 and15. Why do these predictions not result in whole numbers of trout?

4.31 Exercise 4.1 gives the scores of eight subjects on two tests of differing aspects of spatial reasoning:ability to solve geometric problems, x and ability to judge distances, y.

(a) Find the regression line for this data. What does the equation of the line imply about therelationship between these two scores?

(b) Use the regression equation to predict the ability to judge distances score for a subject withgeometric problem solving score of 100. Why is it not a good idea to make a prediction forx =120?

4.32 A subset of the data from the baby tuatara first referred to in Exercise 2.6 is shown below. Re-member that the eggs were incubated in the lab at 18, 21 or 22 degrees Celsius. After hatching, theyoung tuatara had their development monitored. The measurement is SVL = snout-vent length(mm) at age 10 months.

36

◦C 18 18 18 18 21 21 21 21 22 22 22 22SVL 80 81 73 81 91 83 83 89 81 84 76 88

(a) Draw a scatterplot of this data and find the Pearson’s correlation coefficient and the regressionline for this data.

(b) What do the results in (a) imply about the relationship between incubation temperature andSVL?

(c) Use the regression line to predict SVLs for incubation temperatures of 18, 20, 21 and 22 ◦C.

4.33 It has been suggested that much delinquent behaviour is related to undetected hearing impairment.A group of ten delinquent adolescents who had all come to the attention of the courts, were scoredon a psychometric index of delinquency and were also tested for hearing impairment (if any). Theresults were:

Degree of hearing impairment (%), x 0 10 15 20 20 25 25 25 30 40Delinquency index score, y 65 75 91 86 87 91 90 85 78 96

(a) Plot this data on a scatterplot and describe any unusual features.(b) Calculate the least squares regression line and draw it on your plot.(c) Predict delinquency index scores for adolescents with 35% and 50% hearing impairments.

Which of these predictions have you least confidence in and why?

The final set of exercises is for calculation of residuals.

4.34 Following Exercise 4.20:

(a) Calculate the residuals for each strontium level.(b) Draw a plot of residuals versus x for this data. What does it suggest about the assumptions

required for a regression to be a reliable predictor? Give reasons for your answer.(c) How confident are you of the validity of your predictions in Exercise 4.20? Give reasons for

your answer.

4.35 Following Exercise 4.10:

(a) Calculate the residuals for each of the ten subjects, and graph the spatial ability scores againstthese residuals.

(b) Does this graph indicate whether or not the assumptions required for a regression to be validare holding? Give reasons for your answer.

(c) Explain in words what a residual represents.

4.36 Data is collected on the weekly number of calls made by a phone-salesperson over an eight-weekperiod, and the number of sales they actually make in each of those weeks. The training manualspecifies staff should be making sales on at least 25% of their calls. The data are:

Week 1 2 3 4 5 6 7 8

Calls 66 43 57 32 18 59 61 32Sales 20 15 18 12 2 21 18 8

(a) Draw a scatterplot of these data and comment on the suitability of linear regression to modelthe relationship.

(b) Calculate the regression coefficients and write down the estimated prediction equation.(c) Calculate the predictions for x = 20 and x = 60 and use these to superimpose the regression

line on your plot.(d) Calculate the residuals for the sales for weeks 5 and 8, and interpret these numbers.

4.37 Calculate the residuals for each of the twenty geckos in Exercises 4.14 and 4.24, and graph themetabolic rates in trial one against these residuals. Does this graph indicate whether or not theassumptions required for a regression to be valid are holding? Give reasons for your answer.

4.38 For the tuatara in Exercise 4.32:

37

(a) Find the residuals for each piece of data.(b) Graph the residuals versus incubation temperature.(c) What does this graph suggest about your regression?

4.39 Using the results from Exercise 4.33, calculate the residuals for each subject. Plot these residualsagainst the explanatory variable, degree of hearing impairment. Describe any unusual features ofthis plot. How confident are you of the predictions made in Exercise 4.33? Why?

4.40 Calculate the residuals for all observations in Exercises 4.4 and 4.29 and plot against compliancescores. What does this plot suggest about the validity of your regression in Exercise 4.29?

Solutions

4.1 The scatterplot shows that the data is not linear. This suggests that even though both tests aresupposed to measure spatial reasoning, they must actually be measuring different things.

90 95 100 105 110 115

2021

2223

2425

Scatterplot of Spatial Reasoning Tests

Test I score

Tes

t II s

core

4.2 There is a positive linear relationship between oxygen level and number of trout. As oxygen levelincreases, so does the number of trout.

15 20 25

2025

3035

40

Scatterplot of Lake Health

x, oxygen level

y, n

umbe

r of

trou

t

(2)

4.3 There are two possible outliers (15.9, 4) and (4.5, 8.6). If these are ignored, the scatterplot shows apositive linear relationship between x and y, so most subjects that had a fast reaction time when

38

shown a pleasant image, also had a fast reaction time when shown an unpleasant image.

2 4 6 8 10 12 14 16

23

45

67

8

Scatterplot of Reaction Times

x, pleasant image

y, u

nple

asan

t im

age

(2)

4.4 The relationship is positive, as one variable increases, the other tends to increase as well. Thereappears to be a non-linear relationship. For x < 120, y doesn’t increase greatly, but for x ≥ 120,y starts to increase very fast.

90 100 110 120 130

4050

6070

8090

Scatterplot of the Effect of Group Pressure

x, orders from authority

y, w

illin

gnes

s to

be

invo

lved

4.5 There doesn’t seem to be a relationship between the two variables. A possible relationship couldbe negative - as the amount of land acquired increases, the price/acre decreases, but we would needmore data to confirm this. If the data from the Wairarapa is omitted, the price and amount arefairly constant for the remaining data.

39

500 1000 1500 2000 2500 3000 3500

68

1012

14

Scatterplot of Maori Land Purchases

Amount acquired (x1000 acres)

Pric

e/ac

re (

penc

e)

4.6 (a)

5 10 15 20 25 30

510

1520

2530

3540

Scatterplot of Anxiety and Self−esteem

Anxiety level

Sel

f−es

teem

(2)

(b) Although the data are measurements, the relationship is not linear, so Spearman’s is preferred.(c) r = 0.833.(d) Anxiety, x Self-esteem, y r(x) r(y) d d2

2 5 1 1 0 05 20 2 2 0 06 22 3 3 0 010 30 6 6.5 -0.5 0.2510 26 6 4 2 417 35 8 8 0 030 40 9 9 0 010 30 6 6.5 -0.5 0.258 27 4 5 -1 1

∑

d2

i = 5.5 and rs = 1 − 6×5.5720

= 0.954.(e) There is a strong positive (non-linear) relationship between anxiety level and self-esteem.

4.7 (Weakest) 0.0, -0.4, 0.5, 0.6, -0.8 (Strongest).

4.8 (a) The scatterplot (below) has a random pattern of points, so r ≈ 0.(b) looking at the sample of circled points on the scatterplot, we get a positive linear pattern withr close to 1.

40

1 2 3 4 5 6 7

34

56

7

x

y

O O

O O

O

O

4.9 (a) Observations need to be made on each individual for both variables. One person cannot have aman’s wage, and a woman’s wage.(b) −1 ≤ r ≤ 1, so r can’t be 1.21.(c) You can’t find a correlation with a categorical variable (ethnicity).(d) r = 0.18 is positive, not negative. Occupation is categorical.(e) r = −0.94 is very close to −1, so this a strong correlation.

4.10 (a) r = 0.925

6.5 7.0 7.5 8.0 8.5 9.0

4550

5560

6570

Scatterplot of Recollection

Spatial ability

Acc

urac

y of

rec

olle

ctio

n

(2)

(b) The correlation coefficient is positive and close to one. This means there is a strong, positivelinear relationship between spatial ability and accuracy of recollection.(c) Spatial, x Accuracy, y r(x) r(y) d d

2

8.9 60 8 8 0 07.9 56 6 6 0 07.6 53 4.5 4.5 0 09.2 71 10 10 0 09.1 68 9 9 0 06.3 50 1 3 -2 48.6 59 7 7 0 06.4 47 2 2 0 07.6 53 4.5 4.5 0 06.5 42 3 1 2 4

∑

d2

i = 8 and rs = 1 − 6×8

990= 0.952 (3dp).

41

Spearman’s is appropriate because while we are unsure of the actual values, we are confident thatthe data are ordinal, and so rankings based on these scores will be accurate.

4.11 Spearman’s is appropriate here as number of episodes is discrete.Dopamine, x Episodes, y r(x) r(y) d d2

6.5 1 2 1.5 0.5 0.258.3 4 7 7.5 -0.5 0.257.8 3 5 5.5 -0.5 0.259.7 4 8 7.5 0.5 0.258.1 3 6 5.5 0.5 0.257.5 2 4 3.5 0.5 0.257.0 2 3 3.5 -0.5 0.256.1 1 1 1.5 -0.5 0.25

∑

d2

i = 2 and rs = 1 − 6×2

504= 0.976.

4.12 (a) Question C Question D r(C) r(D) d d2

2 6 1 5 -4 165 5 5 2.5 2.5 6.253 6 2.5 5 -2.5 6.255 3 5 1 4 167 7 8 7.5 0.5 0.256 5 7 2.5 4.5 20.255 6 5 5 0 0.253 7 2.5 7.5 -5 25

∑

d2

i = 90.25 and rs = 1 − 6×90.25504

= −0.0744.(b) This is a very weak negative relationship. There is virtually no regular pattern between scoreson these two questions.(c) Scores are ordinal only. The score 2 is not twice the score 1, but is more than it, etc.

4.13 (a)

10 15 20 25 30 35

1520

2530

3540

45

Scatterplot of Mentoring

mentor

trai

nee

(b) r = 0.808, there is a strong positive linear relationship between mentor’s score and the trainee’sscore.(c) Mentor, x Trainee, y r(x) r(y) d d2

23 30 6 7 -1 137 44 8 8 0 07 21 1 3 -2 418 17 5 2 3 910 25 2 5 -3 915 22 4 4 0 013 16 3 1 2 429 28 7 6 1 1

∑

d2i = 28 and rs = 1 − 6×28

504= 0.667.

(d) It is unlikely that these scores are genuine measurements, they are much more likely to be

42

ordinal data, so Spearman’s is more appropriate.

4.14 (a)

0.06 0.08 0.10 0.12 0.14 0.16 0.18

0.06

0.08

0.10

0.12

0.14

Scatterplot of Gecko Metabolic Rate

Vol1 (mL/gram)

Vol

2 (m

L/gr

am)

(b) r = 0.793.

4.15 Duration, x Severity, y r(x) r(y) d d2

6 2 6 3 3 91 2 1 3 -2 4

3.5 2 5 3 2 41.5 2 2 3 -1 12.5 3 3 7 -4 1618 4 9 9 0 011 3 8 7 1 18 3 7 7 0 03 2 4 3 1 1

∑

d2i = 36 and rs = 1 − 6×36

720= 0.7. This implies that there is a strong positive relationship

between duration and severity of Parkinson’s disease, i.e. severity increases over time.

4.16 Pearson’s r = 0.299.Anxiety, x Avoidance, y r(x) r(y) d d2

2.17 0.89 3 1 2 40.89 1.11 1 2 -1 13.11 2.5 7.5 6 1.5 2.253.11 2.72 7.5 8 -0.5 0.252.28 3.17 4 9 -5 253.00 2.61 6 7 -1 10.94 2.39 2 4 -2 42.89 2.44 5 5 0 03.94 1.78 9 3 6 36

∑

d2i = 73.5 and rs = 1 − 6×73.5

720= 0.3875.

The data is widely scattered and hard to see whether there is a linear relationship, so Spearman’sis more appropriate.

43

1.0 1.5 2.0 2.5 3.0 3.5 4.0

1.0

1.5

2.0

2.5

3.0

Scatterplot of Avoidance vs. Anxiety

Anxiety, x

Avo

idan

ce, y

4.17 r = 0.192. There is a weak positive relationship between social dominance orientation score andsupport for biculturalism in principle.

4.18 (a)

−0.2 −0.1 0.0 0.1 0.2

−0.

050.

000.

050.

100.

15

Scatterplot of Stock Exchange Returns

NZSX10

AS

X

(b) r = 0.558. There is a moderate positive relationship between the NZSX and ASX. If one marketvalue goes up, the other is also likely to rise.

4.19 Shanghai and Kaohsiung switch rankings from 2001 to 2002, but all other rankings are the same.∑

d2i = 2 and rs = 1 − 6×2

120= 0.9.

44

4.20 (a)

4.0 4.5 5.0 5.5 6.0

1416

1820

22

Scatterplot of Heavy Metals in Baltic Sea

Strontium (% total weight)

Lead

(%

tota

l wei

ght)

(b) Slope b = 2.26 and intercept a = 6.18. Equation of regression line is y = 6.18 + 2.26x.(c) When x = 4.0, y = 15.2 and when x = 6.0, y = 19.71.(d) We should not predict for x = 7 as it is outside the original range of data.

4.21 (a)

1 2 3 4 5 6 7

200

300

400

500

Scatterplot of Residential House Sales

no. of bedrooms

Pric

e ($

000)

(b) r = 0.848. (c) Slope b = 49.00 and intercept a = 74.82. Equation of regression line isy = 74.82 + 49.00x.(d) x = 1 gives y = 123.82, x = 2 gives y = 172.83, x = 3 gives y = 221.83, x = 4 gives y = 270.83,x = 5 gives y = 319.84, x = 6 gives y = 368.84, and x = 7 gives y = 417.85.

4.22 (a) Slope b = 7.35 and intercept a = −1.48. Equation of regression line is y = −1.48 + 7.35x.

45

6.5 7.0 7.5 8.0 8.5 9.0

4550

5560

6570

Scatterplot of Recollection

Spatial ability

Acc

urac

y of

rec

olle

ctio

n

(2)

(b) r2 = 0.8555(c) When x = 6.8, y = −1.48 + 7.35 × 6.8 = 48.5.

4.23 (a) Slope b = 0.72 and intercept a = 11.77. Equation of regression line is y = 11.77 + 0.72x.(b) When x = 25, y = 11.77 + 0.72 × 25 = 29.67.(c) It does not provide a reliable score as x = 45 is just outside the data range.

4.24 (a) Slope b = 0.581 and intercept a = 0.029.(b) When x = 0.070, y = 0.029 + 0.581 × 0.070 = 0.07 (2dp).

4.25 (a)

0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

2224

2628

Scatterplot of Advertising and Profit

advertising ($000)

prof

it ($

000)

(b) Slope b = 21.47 and intercept a = 11.13. Equation of regression line is y = 11.13 + 21.47x.

4.26 (a) Slope b = 0.206 and intercept a = 2.861. Equation of regression line is y = 2.861 + 0.206x.(b) When x = 2.5, predict support for biculturalism score y = 3.38. When x = 3.0, predicty = 3.48.

4.27 (a) r = 0.816, r2 = 0.665.

46

16000 18000 20000 22000

2530

3540

45

Scatterplot of Income and White Collar Workers

income ($)

whi

te c

olla

r w

orke

rs (

%)

(b) r2 is the proportion of variability of % white collar workers than is explained by income.(c) Southland (17300, 25.4) is a possible outlier. Wellington and Auckland have much larger incomevalues than the other regions.(d) Slope is b = 0.00224 and intercept a = −5.831. Equation of regression line is y = −5.831 +0.00224x.(e) y = −5.831 + 0.00224× 17500 = 33.37%.

4.28 (a) r = 0.357. b = 0.192, a = 3.000, and the regression line equation is y = 3.000 + 0.192x.(b) r2 = 0.128, i.e. 12.8%.(c) x = 5 gives y = 3.958, x = 10 gives y = 4.917, x = 15 gives y = 5.875.

4.29 a = −78.610, b = 1.267.

90 100 110 120 130

4050

6070

8090

Scatterplot of the Effect of Group Pressure

x, orders from authority

y, w

illin

gnes

s to

be

invo

lved

All of the data lie close to the regression line but between 90 and 120, the regression equationoverestimates, and for other values underestimates. When checking the residuals of our regression,there would be a clear pattern. The data is not linear.

4.30 Slope b = 1.633 and intercept a = −1.417. Equation of regression line is y = −1.417 + 1.633x.When x = 20, y = −1.417 + 1.633× 20 = 31.250. When x = 15, y = −1.417 + 1.633× 15 = 23.083.These predictions don’t result in whole numbers because these are predicted trout numbers and

47

are essentially averages for all likely situations when x = 20 and x = 15.

4.31 (a) Equation of regression line is y = 25.424 − 0.031x. The slope is negative, so as one scoreincreases, the other tends to decrease.(b) When x = 100, y = 25.424− 0.031× 100 = 22.340. It is not a good idea to predict for x = 120because this is outside of the data range.

4.32 (a) r = 0.437. Regression line is y = 56.69 + 1.27x.

18 19 20 21 22

7580

8590

Scatterplot of Young Tuatara

temperature

SV

L (m

m)

(2)

(2)

(b) There is a moderate positive relationship between incubation temperature and SVL, i.e. as oneincreases, the other tends to increase.(c) Predicted SVLs for incubation temperatures of 18, 20, 21, and 22◦C are 79.54, 82.08, 83.35,and 84.62 respectively.

4.33 (a) It looks like two different regression lines could be fitted to the data. If the values at x = 0, 10and 40 were considered outliers you would get a negatively sloped regression line through the restof the data, but in general as hearing loss increases so does delinquency index.

0 10 20 30 40

6570

7580

8590

95

Scatterplot of Delinquency

hearing impairment (%)

Del

inqu

ency

inde

x sc

ore

(b) y = 71.280 + 0.625x.(c) For x = 35 we predict y = 93.147, and for x = 50 we predict y = 102.518. We don’t haveconfidence in the latter prediction because it’s outside of the data range.

48

4.34 (a) The residuals are: -1.199, -1.582, 2.673, -1.513, -1.709, 2.855, -0.327, 0.801.(b) There appears to be a relatively even band of points above and below zero (all within ±3) andthere is no obvious pattern. This suggests the residuals are symmetric about zero, with constantvariance.

4.0 4.5 5.0 5.5 6.0

−1

01

23

Residual Plot of Heavy Metals in Baltic Sea

Strontium (% total weight)

Res

idua

ls

(c) From (b), we can be confident of our predictions as it supports that the data is linear and theregression is valid.

4.35 (a) The residuals are: -3.908, -0.561, -1.357, 4.888, 2.622, 5.194, -2.704, 1.459, -1.357, -4.275.

6.5 7.0 7.5 8.0 8.5 9.0

−4

−2

02

4

Residual Plot for Recollection

Spatial Ability

Res

idua

ls

(b) The constant variance assumption appears to hold because there are constant bands of residualscentered around 0, i.e. no funneling. There does appear to be a pattern in the residuals as the onlypositive residuals are near the extreme values of spatial ability. Regression lines are sensitive tooutliers which is a possible cause of this.(c) A residual is the difference between the observed value and the value predicted by the regressionline. It is the vertical distance between the line and the point.

4.36 (a) The scatterplot looks to have a strong positive linear trend and both variables are quantitative,so linear regression seems appropriate.

49

20 30 40 50 60

510

1520

Scatterplot for Phone−sales

Number of Calls

Sal

es

(b) Slope b = 0.362 and intercept a = −2.398. Equation of regression line is y = −2.398 + 0.362x(c) y = 4.840 and 19.317 respectively. These are indicated on the graph by black circles.(d) The residual for week 5 is -2.117, and for week 8 is -1.183. For both weeks, the regressionequation over-predicts the number of sales.

4.37 The residuals are: 0.015, -0.007, 0.002, -0.008, -0.013, 0.0115, -0.002, 0.003, 0.001, -0.007, 0.004,-0.048, 0.032, -0.006, 0.001, 0.003, -0.001, -0.014, 0.018, 0.015.

0.06 0.08 0.10 0.12 0.14 0.16 0.18

−0.

04−

0.02

0.00

0.02

Residual Plot for Geckos

metabolic rate

resi

dual

s

There appears to be funneling, which would mean that there is not constant variance, and thisimplies that our regression is not valid. The point (0.147,−0.048) looks like an outlier.

4.38 (a) The residuals are: 0.46, 1.46, -6.54, 1.46, 7.65, -0.35, -0.35, 5.65, -3.62, -0.62, -8.62, 3.38.

50

(b)

18 19 20 21 22

−5

05

Residual Plot for Tuatara

temperature

resi

dual

s

(2)

(2)

(c) Graph of x versus residuals suggests possible outlier at (18,−6.54). If this is so, then there ispossible funneling in the graph indicating non-constant variance. Also, whether or not this pointis included, there is a clear curve beginning at (18,−6.54) going up to (21, 7.65) and then down to(22,−8.62). so there are reasons to doubt whether the assumptions required for regression to bevalid are holding.

4.39 The residuals are: -6.280, -2.528, 10.349, 2.225, 3.225, 4.101, 3.101, -1.899, -12.023, -0.271.

0 10 20 30 40

−10

−5

05

10

Residual Plot for Delinquency

hearing impairment

resi

dual

s

(15, 10.349) and (30,−12.023) are probably outliers which is consistent with Exercise 4.33. Only xvalues between 20 and 25 have positive residuals, so we might doubt our predictions because ourresiduals do not look random. Otherwise even bands above and below zero (roughly between ±12)and no funneling so it’s possible the regression is valid.

4.40 The residuals are: 8.345, -10.469, -3.527, -14.282, 12.520, 11.450, 3.810, 4.311, -12.608, 0.450. Thereis a clear curved pattern in the residuals, so we would doubt our predictions in Exercise 4.29.

51

90 100 110 120 130

−15

−10

−5

05

10

Residual Plot for Effect of Group Pressure

compliance

resi

dual

s

52

Chapter 5

Exercises

5.1 Which of the following are independent events, or, in the case of some real-life examples, almostindependent?

(a) The size of a student loan is over $10,000 and the student is studying for a medical degree.(b) New Zealand student, Nick Willis makes the finals in the 1500m at the 2004 Olympics and the

Evers-Swindell sisters win gold in the Olympic 2000m double sculls.(c) It is raining in Auckland and raining in Hokitika.(d) The length of sentence in a criminal conviction is over 12 months and the crime is grievous

bodily harm.(e) Having prostate cancer and returning a positive test for prostate cancer.(f) An area has a high level of overcrowding and the incidence of TB in that area is high.(g) The stoat population in an area is high and the size of the kiwi population in that area is low.(h) Global warming is increasing and the sheep and cow populations in New Zealand are increasing.(i) Oil prices are high and the situation in the Middle East is unstable.

5.2 Which of the following are mutually exclusive events?

(a) The event that Nick Willis won the 1500m at Athens, 2004 and the event that he came second.(b) The event that Nick Willis won the 1500m at Athens, 2004 and the event that he put in a

personal best time in that event.(c) The event of passing a 100-level statistics course and that of having to retake the course.(d) The event of passing a 100-level statistics course and that of passing a 200-level statistics

course.(e) The event of having a student loan of less than $20,000 and that of having one more than

$25,000.(f) The event that a rash turns out to be measles and the event that it turns out to be chicken

pox.

This second set of exercises gives practice with drawing probability trees, and using them to find proba-bilities.

5.3 Toxemia is a potentially dangerous condition of pregnancy which is more likely to occur in diabeticmothers than non-diabetics. About 2% of the general female population are diabetic and of these,25% are likely to develop toxemia during pregnancy. In the non-diabetic female population, 4.6%develop toxemia during pregnancy. Consider a randomly selected pregnant woman.

(a) Draw a tree diagram representing all possible outcomes for this woman.(b) Attach to the diagram the probabilities for each branch and outcome.(c) What is the probability a diabetic woman completes her pregnancy without toxemia?(d) What is the probability that a pregnant woman develops toxemia?

5.4 A survey was conducted of bail decisions in the district court over a three-month period. In thatperiod all defendants fell into three groups: Maori (M), Pacific Islands (P) or European (E). Fiftypercent were European and of these 80% were granted bail (B). Thirty-five percent were Maori andof these 40% were granted bail. In the case of Pacific Islands defendants, 60% were not grantedbail.

(a) Draw a tree diagram showing all the possible outcomes for a randomly selected defendant inthis time period.

(b) Write the appropriate probabilities on each branch of the tree and calculate the probabilitiesof all outcomes and write them on the right-hand side of the tree.

(c) Calculate the probability that a randomly selected defendant will have been granted bail.

5.5 A survey was conducted in two areas of greater Wellington, Mana (M) and Lower Hutt (L) in orderto evaluate policing needs and levels of victimisation in these areas. Forty percent of the surveyarea population live in Mana and of these, 18% had experienced offences against their person (O).In Lower Hutt, 13% of residents had been the victim of offences against their person.

53

(a) Draw a tree diagram showing all of the possible outcomes for a randomly selected person fromthe Mana-Lower Hutt region.

(b) Write the appropriate probabilities on each branch of the tree and calculate the probabilitiesof all outcomes.

(c) Calculate the probability that a randomly selected person from the region will have been thevictim of an offence against their person.

5.6 An experiment with rats involves a maze with only two routes available. The choice is whetherto swim across a water hazard to where a reward of food can be seen (WET), or to take a longerdry route (DRY) where there is no visible reward. Sixty percent of the rats have recently been fed(FULL) and 40% have not been fed for a day (HUNGRY). Of the full rats 60% chose the wet routeand 20% of the hungry rats chose the dry route.

(a) Draw a tree diagram showing all the possible outcomes for a randomly selected rat with theappropriate probabilities on each branch of the tree.

(b) Calculate the probabilities of all outcomes and write them on the right-hand side of the tree.(c) Calculate the probability that a randomly selected rat will choose the wet route.

5.7 A group of young men from Hawke’s Bay, who had all at one time been on the CYPS register ofchildren at risk, were followed up at age 19. Forty percent of these young men were found to begang members (G) and 15% were gang prospects (P). The remainder had no gang affiliations (N).The history of these young men in the criminal justice system was investigated and the maximumpenalty they had incurred to date was recorded. Each fell into one of three categories: custodialsentence, no convictions, fine and/or community service. Of the gang members, 8% had had noconvictions, 52% had incurred a fine and/or community service while the remainder had alreadyserved a custodial sentence. Sixty percent of the gang prospects had had a fine and/or communityservice and 30% had served a custodial sentence. Of those with no gang affiliations 50% had noconvictions and 10% had custodial sentences.

(a) Consider a randomly selected young man from this group and draw a tree diagram representingall possible outcomes for this young man. Attach to the diagram the appropriate probabilitiesfor each branch and outcome.

(b) What is the probability that such a randomly selected 19-year-old male has no gang affiliationsand has served a custodial sentence?

(c) What is the probability that the randomly selected 19-year-old has already served a custodialsentence?

5.8 Exercise 3.13 gave data on the cost of power generation in New Zealand in the mid-1980s. Considera randomly selected power station of that era. The probability that the power station is a hydro-station is 21/37. If hydro-powered, the probability of it producing power at less than 1 cent perkW is 20/21, but if not hydro-powered, the probability of it producing power at less than 1 centper kW is only 25%. What is the probability that a randomly selected power station producespower at less than 1 cent per kW?

5.9 In New Zealand, 5% of the population belong to one of the Pacific Island groups (P), 15% are Maori(M) and the rest are European or Other (EO). In the 2001 census, a family was defined as oneor two parent(s) and their children living in the same household (C) or a same household couplewithout children (NC). Different groups in New Zealand have very different proportions of thesetwo family types. Of the Pacific Island group, the proportion of families living as a couple withoutchildren (NC) is a low 8%, for the Maori population it is 12%, while amongst the European andOther group it is much higher at 28.5%.

(a) Consider a randomly selected family in New Zealand and draw a tree diagram of all possibleoutcomes for that family.

(b) What is the probability that a randomly selected family consists of a couple with no children?What percentage is this?

(Data based on figures published by Statistics New Zealand.)

5.10 In a study of the effect of health levels on the ability of an individual to save for retirement a largesample was taken from the population of over-40-year-olds. In this group 45% were classified as

54

having a low level of health (L) and the remainder had a satisfactory or high health level (H). Ofthose with satisfactory to high health levels, 80% were saving for retirement (S) in an adequateway. Of those with low health levels 30% were not saving adequately (N). Consider a randomlyselected person from this sample.

(a) Draw a tree diagram representing all possible outcomes for the person’s health level and savingbehaviour with appropriate probabilities on each branch.

(b) Calculate the probabilities of all possible outcomes.(c) What is the probability that a person has a low health level and is managing to save adequately

for retirement?(d) What is the probability that a person is not saving adequately for retirement?

5.11 When real estate is listed for sale, it can either be with a sole agent, or listed by multiple real estateagents. Approximately 95% of listings are sole listings. When the house does not sell in the firstmonth, the probability of it selling in the second month if it is a sole listing is 40%, whereas if ithas multiple listings, the probability is 20%.

(a) Draw a probability tree describing this situation.(b) For a randomly selected house that has not sold within the first month, what is the probability

that it sells in the second month?

5.12 In a South Auckland community, 20% of children at primary school were identified as failing tolearn in a way commensurate with development (F). Of these children, 30% were subsequentlyfound to have a psychological problem such as dyslexia, and 50% were found to have a physicalproblem such as “glue” ear. The remainder had no discernible cause for their failure to learn. Thechildren who were not failing to learn were also tested and a surprising 2% had a physical problemwhile 5% had a psychological problem.

(a) Consider a randomly selected child from this community and draw a tree diagram representingall possible outcomes for this child.

(b) Calculate the probabilities of all of the possible outcomes associated with this tree diagram.What is the probability that such a randomly selected child is not identified as having anyimpairment to learning and has a physical problem?

(c) What is the probability that the randomly selected child has a psychological problem? Whatpercentage of children is this?

5.13 Males (M) are more likely to be affected with schizophrenia than females (F) and have a chanceof being affected with schizophrenia of 0.3%, while females have a chance of 0.2 in 100. A drugtreatment (D) used only in schizophrenia is more likely to be used with males with whom it ismore successful, and the probability that it is prescribed for them is 0.7. The probability that it isprescribed for a female sufferer is 0.4.

(a) In a community of equal numbers of males and females, consider a randomly selected member ofthis community and draw a tree diagram representing all possible outcomes for this individual’sgender, mental health and treatment.

(b) Attach to the diagram the probabilities for each branch and outcome.(c) What is the probability that such a randomly selected individual is male with schizophrenia

and not receiving the drug treatment?(d) What is the probability that a randomly selected individual is receiving the drug treatment?(e) Draw up a probability distribution in this situation, i.e. a list of outcomes and their associated

probabilities.

5.14 In a particular part of the North Island in New Zealand two varieties of a Hebe plant are found,macroura (M) and atkinsonii (A). 70% of Hebes in this region are of variety M. Plants are classifiedas M or A in the field by looking at leaf length and this method is not perfect. Variety M iscorrectly identified (C) by this method 60% of the time, and there is a 6% chance of an A plantbeing incorrectly identified using this method.

(a) Draw a tree diagram showing the possible outcomes for a randomly selected Hebe from thisarea and calculate the probabilities for all possible outcomes.

55

(b) Calculate the probability that a randomly selected Hebe plant is correctly identified by thismethod.

(c) Draw up a probability distribution in this situation, i.e. a list of outcomes and their associatedprobabilities.

5.15 A sociologist is studying the relationship within families of the highest educational level achieved.Pairs of brothers, where both are aged between 30 and 50, were investigated. In this group 5% ofthe elder brothers had primary school education only while 80% had a secondary education andthe remainder had some tertiary education. Of those elder brothers who had primary educationonly, 80% had a younger brother who also had primary school education only, while none had ayounger brother with tertiary education. Of those elder brothers with secondary education only,50% had younger brothers who also had secondary level education and 10% had younger brotherswith primary education only. Forty percent of those elder brothers with tertiary education hadyounger brothers who also had tertiary level education while the remainder had brothers withsecondary level education.

(a) Draw a tree diagram showing all possible outcomes for a randomly selected pair of brothers inthe study age group.

(b) Calculate the probabilities of all possible outcomes.(c) Calculate the probability that a randomly selected pair of brothers in the study age group have

the same highest level of education.(d) Calculate the probability that a randomly selected pair of brothers in the study age group have

a different highest level of education(e) Calculate the probability that a randomly selected pair of brothers in the study age group have

the younger brother with a higher level of education than the elder.(f) Draw up a probability distribution in this situation, i.e. a list of outcomes and their associated

probabilities.

This final set of exercises gives practice using Bayes’ rule. Most of these problems require you to havecompleted one of the earlier exercises.

5.16 In Exercise 5.3, use the tree diagram to find the proportion of those women developing toxemiathat are non-diabetic?

5.17 In Exercise 5.4, what proportion of those granted bail in the district court are Maori?

5.18 In Exercise 5.5, what is the probability that a person who had been the victim of an offence againsttheir person comes from Mana?

5.19 In Exercise 5.7, referring to young men in Hawke’s Bay who had at one time been on the CYPSregister, what proportion of those young men who had incurred a fine and/or community servicewere gang members?

5.20 In Exercise 5.8, what is the probability that a power station that produces power at less than 1cent per kW is a hydro-station?

5.21 In Exercise 5.9, what percentage of families with no children in New Zealand are Pacific Islandfamilies?

5.22 Twenty percent of products in a small supermarket have very short lifespans, i.e. they go off quickly.Of these short lifespan products, there is a 30% chance that a randomly selected item is past itsuse-by date, whereas with the longer lasting products, there is only a 12% chance that it is past itsuse-by date. What is the probability that a product past its use-by date is a long-lasting product?

5.23 Following from Exercise 5.11, if a randomly selected house was sold in the second month, what isthe probability that it was a sole listed property?

5.24 In Exercise 5.12, in the South Auckland community studied, what percentage of children withpsychological problems are failing to learn?

56

5.25 Consider a hypothetical town with population 10,000 in which a man is suspected of a crime. Hisblood matches a stain found at the scene some time after the crime was committed. 1% of thepopulation will have matching blood. Because in this case the stain is somewhat degraded, theprobability of getting a match even if the blood is identical is 0.95.

(a) Draw a tree of the different possibilities of the man being guilty or not and his blood matchingor not.

(b) Find the probability of a match.(c) Find the probability that this man is guilty of the crime and his blood matches.(d) If there is a match, find the probability that the man is guilty. (Use Bayes’ rule.)

Solutions

5.1 (a) This certainly wasn’t independent in the mid-90s when general academic fees were small, how-ever, with the rise of fees for general courses, these events are probably close to being independent.(b) Independent events, unlikely to affect each other’s performance except in terms of overallmorale.(c) Loosely independent, associated with overall weather in New Zealand.(d) Not independent, if you commit grievous bodily harm, you are likely to get a sentence longerthan a year.(e) Not independent, it is less likely to get a false positive.(f) Not independent, TB is related to poor housing.(g) Not independent, the size of the kiwi population is directly (inversely) related to the predatorpopulation.(h) Not independent, global warming and stock population are likely to be related, due to therelease of methane gas.(i) Not independent, the Middle East is a major source of oil.

5.2 (a) Mutually exclusive, you can’t come first and second in a race.(b) Not mutually exclusive, you can win and have a personal best in one race.(c) Mutually exclusive, if you pass, you will not have to retake.(d) Not mutually exclusive, to pass a second year course, you usually need to pass a first yearcourse first.(e) Mutually exclusive, your student loan cannot be less than $20k and more than a larger value.(f) Mutually exclusive, it cannot be both measles and chicken pox.

5.3 (a) and (b) Define D = {woman diabetic} and T = {toxemia develops}.

D

D

T

T

T

T

0.02

0.98

0.25

0.75

0.046

0.954

P (D and T ) = 0.02 × 0.25 = 0.00500

P (D and T ) = 0.02 × 0.75 = 0.01500

P (D and T ) = 0.98 × 0.046 = 0.04508

P (D and T ) = 0.98 × 0.954 = 0.93492

(c) P (diabetic woman completes her pregnancy without toxemia) = 0.75 = 75%.(d) P (T ) = 0.005 + 0.04508 = 5.008%.

57

5.4 (a) and (b) Define M = {Maori}, P = {Pacific Islands}, E = {European} and B = {granted bail}.

M

P

E

B

B

B

B

B

B

0.35

0.15

0.5

0.4

0.6

0.4

0.6

0.8

0.2

P (M and B) = 0.35 × 0.4 = 0.14

P (M and B) = 0.35 × 0.6 = 0.21

P (P and B) = 0.15 × 0.4 = 0.06

P (P and B) = 0.15 × 0.6 = 0.09

P (E and B) = 0.5 × 0.8 = 0.4

P (E and B) = 0.5 × 0.2 = 0.1

(c) P (B) = 0.14 + 0.06 + 0.40 = 0.6 = 60%.

5.5 (a) and (b) Define M = {Mana} and O = {offended against}.

M

M

O

O

O

O

0.4

0.6

0.18

0.82

0.13

0.87

P (M and O) = 0.4 × 0.18 = 0.072

P (M and O) = 0.4 × 0.82 = 0.328

P (M and O) = 0.6 × 0.13 = 0.078

P (M and O) = 0.6 × 0.87 = 0.522

(c) P (O) = 0.072 + 0.078 = 0.15 = 15%.

5.6 (a) and (b) Define F = {full} and W = {chooses wet}.

F

F

W

W

W

W

0.6

0.4

0.6

0.4

0.8

0.2

P (F and W ) = 0.6 × 0.6 = 0.36

P (F and W ) = 0.6 × 0.4 = 0.24

P (F and W ) = 0.4 × 0.8 = 0.32

P (F and W ) = 0.4 × 0.2 = 0.08

(c) P (W ) = P (F and W ) + P (F and W ) = 0.36 + 0.32 = 0.68 = 68%.

5.7 (a) Define G = {gang member}, P = {prospect}, N = {no gang affiliation}, NC = {no convictions},

58

F = {fine and/or community service}, and C = {custodial sentence}.

G

P

N

NC

F

C

NC

F

C

NC

F

C

0.4

0.15

0.45

0.08

0.52

0.4

0.1

0.6

0.3

0.5

0.4

0.1

P (G and NC) = 0.4 × 0.08 = 0.032

P (G and F ) = 0.4 × 0.52 = 0.208

P (G and C) = 0.4 × 0.4 = 0.16

P (P and NC) = 0.15 × 0.1 = 0.015

P (P and F ) = 0.15 × 0.6 = 0.09

P (P and C) = 0.15 × 0.3 = 0.045

P (N and NC) = 0.45 × 0.5 = 0.225

P (N and F ) = 0.45 × 0.4 = 0.18

P (N and C) = 0.45 × 0.1 = 0.045

(b) P (N and C) = 0.045 = 4.5%.(c) P (C) = 0.16 + 0.045 + 0.045 = 0.25 = 25%.

5.8 P (produces < 1%) = P (Hydro and produces < 1%) + P (not Hydro and produces < 1%) =21

37× 20

21+ 16

37× 1

4= 20

37+ 4

37= 0.649 = 65%.

5.9 (a) Define P = {Pacific Islands}, M = {Maori}, EO = {European or Other} and C = {parentsand children living in same household}.

P

M

EO

C

C

C

C

C

C

0.05

0.15

0.8

0.08

0.92

0.12

0.88

0.285

0.715

P (P and C) = 0.05 × 0.08 = 0.004

P (P and C) = 0.05 × 0.92 = 0.046

P (M and C) = 0.15 × 0.12 = 0.018

P (M and C) = 0.15 × 0.88 = 0.132

P (EO and C) = 0.8 × 0.285 = 0.228

P (EO and C) = 0.8 × 0.715 = 0.572

(b) P (C) = 0.004 + 0.018 + 0.0228 = 0.25.

5.10 (a) and (b) Define L = {low health} and S = {adequate savings}.

L

L

S

S

S

S

0.45

0.55

0.7

0.3

0.8

0.2

P (L and S) = 0.45 × 0.7 = 0.315

P (L and S) = 0.45 × 0.3 = 0.135

P (L and S) = 0.55 × 0.8 = 0.440

P (L and S) = 0.55 × 0.2 = 0.110

(c) P (L and S) = 0.315 = 31.5% (from tree diagram).(d) P (S) = 0.135 + 0.11 = 0.245 = 24.5%.

59

5.11 (a) Define S = {sole listing} and Q = {sells in the second month}.

S

S

Q

Q

Q

Q

0.95

0.05

0.4

0.6

0.2

0.8

P (S and Q) = 0.95 × 0.4 = 0.38

P (S and Q) = 0.95 × 0.6 = 0.57

P (S and Q) = 0.05 × 0.2 = 0.01

P (S and Q) = 0.05 × 0.8 = 0.04

(b) P (Q) = 0.38 + 0.01 = 0.39 = 39%.

5.12 (a) and (b) Define F = {failing to learn}, Ps = {psychological problem}, Ph = {physical problem}and N = {no problem}.

F

F

Ps

Ph

N

Ps

Ph

N

0.2

0.8

0.3

0.5

0.2

0.05

0.02

0.93

P (F and Ps) = 0.2 × 0.3 = 0.060

P (F and Ph) = 0.2 × 0.5 = 0.100

P (F and N) = 0.2 × 0.2 = 0.040

P (F and Ps) = 0.8 × 0.05 = 0.040

P (F and Ph) = 0.8 × 0.02 = 0.016

P (F and N) = 0.8 × 0.93 = 0.744

(b) P (F and Ph) = 0.016 = 1.6% (from tree diagram).(c) P (Ps) = 0.06 + 0.04 = 0.10 = 10%.

5.13 (a) and (b) Define M = {male} and F = {female}, A = {affected}, and D = {prescribed thedrug}.

M

F

A

A

A

A

D

D

D

D

0.5

0.5

0.003

0.997

0.002

0.998

0.7

0.3

0.4

0.6

P (M, A, D) = 0.5 × 0.003 × 0.7 = 0.00105

P (M, A, D) = 0.5 × 0.003 × 0.3 = 0.00045

P (M, A) = 0.5 × 0.997 = 0.4985

P (F, A, D) = 0.5 × 0.002 × 0.4 = 0.0004

P (F, A, D) = 0.5 × 0.002 × 0.6 = 0.0006

P (F, A) = 0.5 × 0.998 = 0.499

(c) P (M and A and D) = 0.00045 = 0.045% (from tree diagram).(d) P (D) = 0.00105 + 0.0004 = 0.00145 = 0.145%(e) P (M, A, D) = 0.00105, P (M, A, D) = 0.00045, P (M, A) = 0.4985, P (F, A, D) = 0.0004,P (F, A, D) = 0.0006, and P (F, A) = 0.4990.

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5.14 (a) Define M = {variety M} and C = {correctly identified}.

M

M

C

C

C

C

0.7

0.3

0.6

0.4

0.94

0.06

P (M and C) = 0.7 × 0.6 = 0.420

P (M and C) = 0.7 × 0.4 = 0.280

P (M and C) = 0.3 × 0.94 = 0.282

P (M and C) = 0.3 × 0.06 = 0.018

(b) P (C) = 0.42 + 0.282 = 0.702.(c) Outcomes: M identified correctly, with probability = 0.42, M not identified with probability= 0.28, A identified correctly, with probability = 0.282 and A not identified, with probability= 0.018.

5.15 (a) and (b) Defining P1 = {elder has primary education only}, S1 = {elder has secondary educationbut no tertiary}, T1 = {elder has tertiary education}, and P2, S2 and T2 similarly for the youngerbrother.

P1

S1

T1

P2

S2

T2

P2

S2

T2

P2

S2

T2

0.05

0.8

0.15

0.8

0.2

0

0.1

0.5

0.4

0

0.6

0.4

P (P1 and P2) = 0.05 × 0.8 = 0.04

P (P1 and S2) = 0.05 × 0.2 = 0.01

P (P1 and T2) = 0.05 × 0 = 0

P (S1 and P2) = 0.8 × 0.1 = 0.08

P (S1 and S2) = 0.8 × 0.5 = 0.40

P (S1 and T2) = 0.8 × 0.4 = 0.32

P (T1 and P2) = 0.15 × 0 = 0

P (T1 and S2) = 0.15 × 0.6 = 0.09

P (T1 and T2) = 0.15 × 0.4 = 0.06

(c) P (same level) = P (P1 and P2)+P (S1 and S2)+P (T1 and T2) = 0.04+0.4+0.06 = 0.50 = 50%.(d) P (different level of education) = 1 − P (same level) = 0.50.(e) P (younger brother has higher education) = P (P1 and S2) + P (P1 + T2) + P (S1 + T2) = 0.01 +0 + 0.32 = 0.33 = 33%.(f) P (P1 and P2) = 0.04, P (P1 and S2) = 0.01, P (S1 and P2) = 0.08, P (S1 and S2) = 0.4 andP (S1 and T2) = 0.32, P (T1 and S2) = 0.09 and P (T1 and T2) = 0.06.

5.16 Proportion of toxemic women that are non diabetic = P (D given T ) = P (D and T )/P (T ) =0.04508/0.05008 = 0.900 (3dp).

5.17 Proportion of those granted bail that are Maori = P (M given B) = P (M and B)/P (B) = 0.14/0.6 =0.233 (3dp).

5.18 Probability that a person who has been a victim of an offence against their person comes fromMana = P (M given O) = P (M and O)/P (O) = 0.072/0.15 = 0.48.

5.19 Proportion of young men on the CYPS register who had incurred a fine or community service thatwere gang members = P (G given F ) = P (G and F )/P (F ) = 0.208/0.478 = 0.435 (3dp).

5.20 Probability that a power station that produces power at less than 1% is hydro = P (H given <

61

1%) = P (H and < 1%)/P (< 1%) = 20

37/ 24

37= 20

24= 0.833 (3dp).

5.21 Proportion of families with no children in New Zealand that are Pacific families = P (P given NC) =P (P and NC)/P (NC) = 0.004/0.25 = 0.016.

5.22 Define S = {short lifespan} and P = {past use-by date}.

S

S

P

P

P

P

0.2

0.8

0.3

0.7

0.12

0.88

P (S and P ) = 0.2 × 0.3 = 0.060

P (S and P ) = 0.2 × 0.7 = 0.140

P (S and P ) = 0.8 × 0.12 = 0.096

P (S and P ) = 0.8 × 0.88 = 0.704

P (P ) = 0.06 + 0.096 = 0.156. So the probability that a product past its use-by date is a long-lifeproduct = P (L given P ) = P (L and P )/P (P ) = 0.096/0.156 = 0.615 (3dp).

5.23 Probability that a house sold in the second month was a sole listed property = P (S given Q) =P (S and Q)/P (Q) = 0.38/0.39 = 0.974 (3dp).

5.24 Percentage of children with psychological problems that are failing to learn = P (F given Ps) =P (F and Ps)/P (Ps) = 0.06/0.10 = 0.6.

5.25 (a) Define G = {guilty} and B = {blood matches}.

G

G

B

B

B

B

0.0001

0.9999

0.95

0.05

0.01

0.99

P (G and B) = 0.0001 × 0.95 = 0.000095

P (G and B) = 0.0001 × 0.05 = 0.000005

P (G and B) = 0.9999 × 0.01 = 0.009999

P (G and B) = 0.9999 × 0.99 = 0.989901

(b) P (B) = 0.000095 + 0.009999 = 0.010094 = 1.0094%.(c) P (G and B) = 0.000095 = 0.0095%.(d) P (G given B) = P (G and B)/P (B) = 0.000095/0.010094 = 0.0094 = 0.94%.

62

Chapter 6

Exercises

These first exercises are for finding binomial probabilities, and deal with properties of binomial randomvariables.

6.1 The prevalence of bovine tuberculosis (TB in cows) is 5%. Ten cows were randomly selected, eachfrom a different herd and the number of cattle that were found to have bovine TB was counted.Label this variable X .

(a) By using a tree or otherwise find and write down a probability distribution for X .(b) Why was it necessary to choose no more than one cow from a single herd?

6.2 The probability that a randomly selected adolescent asthmatic will have suffered a worsening oftheir asthma since the age of ten is 0.4. In a random sample of 20 adolescent asthmatics thenumber, X , whose asthma has worsened since age ten is counted. Calculate the probabilities ofX = 0, 1, 2, . . . , 20 and list as a probability distribution.

6.3 Following from Exercise 5.3, a subsample of 10 pregnant women is randomly selected. Let X denotethe number who develop toxemia. Take P (toxemia) to 2dp.

(a) What type of distribution does X have? Give reasons for your answer.(b) What are the mean and variance of X?(c) Find P (X = 1) and P (X ≤ 2).

6.4 Using Exercise 5.4, let X be the number of defendants not granted bail out of 20 randomly selecteddefendants in the district court over this period of time.

(a) What type of random variable is X? Give reasons for your answer.(b) What is the probability that X = 7? More than 8?(c) If many samples of 20 defendants from the district court are taken, and in each case the number

who are granted bail, X , is counted, then what is the mean value of X , i.e. the average numberwho are granted bail per sample of 20? What is the variance of X?

6.5 Using Exercise 5.5, let X be the number of people who have been the victim of an offence againsttheir person, out of 20 randomly selected people from the Mana-Lower Hutt region.

(a) What type of random variable is X? Why?(b) What is the probability that X = 5? More than 6?(c) If many samples of 20 people are taken from this region, and in each case the number who

have been victims of offences against their person, X , is counted, then what is the mean valueof X , i.e. the average number who have experienced such crimes per sample of 20? What isthe variance of X?

6.6 From Exercise 5.7, a random sample of 12 such young men is selected. Let X denote the numberwho have already served a custodial sentence.

(a) What are the mean and variance of X?(b) Find P (X = 6) and P (X > 6).

6.7 In June 2003, approximately 45% of New Zealand’s work force were women. In a random sampleof 20 workers:

(a) What is the probability that six are women?(b) What is the probability that there were 12 or more men?

(Source: Statistics New Zealand.)

6.8 In Exercise 5.12, a random sample of 15 children from this community is selected and the number,X , with a psychological problem is counted. Find P (X = 3) and P (X > 3)?

6.9 A random sample of 12 families is selected from the context of Exercise 5.9. Let X be the numberwithout children.

63

(a) What type of distribution does X have? Justify your answer.(b) Find P (X = 5) and P (X < 2).(c) If in this sample, six families have no children, what might you suspect?

6.10 Forty percent of all first admissions to psychiatric facilities in the male 45-54 category in NewZealand have alcohol abuse as the leading cause. Of 15 male first admissions in this 45-54 agegroup what is the probability that fewer than four will be due to alcohol abuse?

6.11 In a study of depressed patients, 60% were classified as introverted and 40% as extroverted person-alities. In a random sample of 15 depressed patients let X = the number of extroverted patients.

(a) What are the conditions required for X to have a binomial distribution?(b) Find P (X = 5), P (X < 5) and P (X > 12).

6.12 The probability of a property being sold by sole agent in the first week of listing is 20%. In arandom sample of eight properties listed by a single real estate agent:

(a) What is the probability that none sell in the first week?(b) What is the probability that at least one sells in the first week?(c) What is the probability that all sell in the first week?

6.13 The shell of the snail Limicolaria martensiana has two colour forms, plain and streaked. It hasbeen theorised that the proportion of streaked shells is 60% in the population. A random sampleof 20 snails is collected and the number of plain shells, X is counted.

(a) Find the probability that exactly nine shells will be plain. Give a reason for your answer.(b) Find P (X ≥ 10).(c) The random sample size is increased to 40. In this new sample there are 29 plain shells. What

can you conclude?

(From Owen, D. (1963), Polymorphism and population density in the African land snail Limicolariamartensiana, Science 140, 666-667.)

6.14 On 31 March 2003, cars made up approximately 70% of New Zealand’s 2,916,734 registered vehicles.Consider a random sample of 20 vehicles.

(a) What is the probability that all of them are cars?(b) How many would you expect to be cars?(c) If you were told that only 10 of them were cars, would this lead you to doubt the randomness

of the sampling process?

(Source: Land Transport Safety Authority.)

6.15 The following are measures of fine motor skill development in 20 toddlers who were exposed toplay involving the development of these skills after the age of 1. The scores are an aggregate ofscores on several tests involving fine motor skills: 7.4, 7.6, 8.1, 8.3, 8.4, 8.6, 9.0, 9.0, 9.8, 9.8, 9.9,10.1, 10.1, 10.3, 10.5, 10.6, 10.8, 10.9, 11.0, 11.0.

(a) Find the proportion of these toddlers whose fine motor skills score is above 10.7.(b) If having a score greater than 10.7 is counted as a “success”, this can be seen as a binomial

experiment. Why?(c) Find the probability that in 20 randomly selected toddlers who were exposed to this play, 14

have a fine motor skills score greater than 10.7.(d) Find the probability that in 20 randomly selected toddlers who were exposed to this play, more

than 14 of them have a fine motor skills score greater than 10.7.

6.16 It is estimated that 30% of short-life products on a supermarket’s shelves are past their use-bydates. Consider a random sample of 15 short-life products from this supermarket.

(a) Justify use of the binomial distribution for the number of items in this sample that are pasttheir use-by dates.

(b) Find the probability that none of the items are past their use-by dates.(c) Find the probability that two or more of the items are past their use-by dates.

64

6.17 The probability of Telecom’s share price going up on a particular day is approximately 55%. AssumeX = the number of increases in 20 consecutive days is binomial, with n = 20 and p = 0.55.

(a) What are the mean and standard deviation of X?(b) What is the probability of getting at least 10 increases in 20 consecutive days?(c) Why might the binomial distribution not be appropriate in this case?

6.18 The chocolate bar producer in Exercise 2.17 distributes the bars in packs of 20. Occasionally, theboxes do not close well because some of the bars are too big. The producer estimates that a bar is“too big” when its weight exceeds 52g, and that if there are more than six of these bars in the box,then the box will not close properly. The probability that a bar is too big is estimated to be 10%.

(a) What are the mean and standard deviation of the number of bars in a box that are too big?(b) What is the probability that a randomly selected box will not close properly?

This second set of questions give further practice finding binomial probabilities, and also practice dealingwith sample proportions.

6.19 A smoking cessation programme has a national success rate of 35%. A sample of n = 15 people whoundertake the programme is selected. Let X represent the number who manage to stop smoking.

(a) Find P (X = 2), P (X > 10) and P (X < 8).(b) If groups of 15 smokers on the programme all over the country try to stop smoking, what is

the mean number of smokers per group that will succeed? What is the variance of the numberthat will succeed?

(c) What is the mean proportion of smokers per group who will succeed? What is the variance ofthe proportion of smokers per group who will succeed?

6.20 Forty-five quadrats were pegged out on the southern coast of Stewart Island. These were observedfor three days in late September and it was recorded whether or not southern skuas were observedin each quadrat (these birds breed prolifically in Stewart Island but rarely in the South Island).The table shows the results for each quadrat, ✓ = observed, and ✗ = not observed.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15✓ ✓ ✓ ✗ ✓ ✗ ✗ ✓ ✓ ✓ ✗ ✓ ✓ ✗ ✓

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30✗ ✓ ✓ ✓ ✓ ✓ ✓ ✗ ✓ ✓ ✓ ✓ ✓ ✓ ✓

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45✓ ✓ ✓ ✗ ✗ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓

(a) In what proportion of quadrats were southern skua observed?(b) In a random sample of 12 quadrats from the total of 45 on the south coast of Stewart Island,

let X = the number of quadrats in which southern skua were observed. Find P (X = 6) andP (6 ≤ X ≤ 10).

(c) If Y = the number of quadrats in these 12 in which southern skua are absent, what is thedistribution of Y ? Find P (Y < 2) and P (Y > 4).

6.21 The red blood cell count (×106 per mL) in a group of ten dogs is as follows: 9.5, 7.5, 7.0, 5.9, 9.5,6.2, 6.5, 8.7, 5.8, 7.7.

(a) What proportion of these dogs have a red blood cell count of less than 6.0?(b) If this group of dogs is representative of the population of dogs in terms of red blood cell count,

what is the probability that in a random sample of 20 dogs less than four of them will have ared blood cell count of 6.0 or more?

6.22 The proportion of subjects in which Wade et al were able to induce a false memory (see Exer-cise 2.22) was 0.4. In a random sample of 11 subjects where X = the number in whom a falsememory is able to be induced find P (X = 5) and P (X ≥ 6).

6.23 In the year ended June 2003, roughly 20% of New Zealand’s $29,278 million exports were to theUnited States. Consider 12 randomly selected export units.

65

(a) What is the probability that three or more of these were sent to the United States?(b) What is the probability that none of them were bound for the United States?(c) What is the probability that seven of them were sent to countries other than the United States?

(Source: Statistics New Zealand.)

Solutions

6.1 (a) X is Bin(10, 0.05), sox 0 1 2 3 4 5 6 . . . . 10

P (X = x) 0.5987 0.3151 0.0746 0.0105 0.0010 0.0001 less than 0.0001

(b) TB is contagious, so for cows (trials) to be independent, need to samples cows from differentherds.

6.2 X is Bin(20, 0.4), sox 0 1 2 3 4 5 6

P (X = x) 0.0000 0.0005 0.0031 0.0123 0.0350 0.0746 0.1244

x 7 8 9 10 11 12 13P (X = x) 0.1659 0.1797 0.1597 0.1171 0.0710 0.0355 0.0146

x 14 15 16 17 18 19 20P (X = x) 0.0049 0.0013 0.0003 0.0000 0.0000 0.0000 0.0000

6.3 (a) p = P (pregnant woman develops toxemia) = P (T ) = 0.05 (to 2dp, from Exercise 5.3); n = 10;X = number in sample who develop toxemia. X has a binomial distribution since there are twooutcomes (develop toxemia or not), the women are independent (unless there are siblings in thesample), the number of trials is fixed at 10 and the probability of developing toxemia is roughlyconstant. X ∼ Bin(10, 0.05).(b) Mean of X = 10 × 0.05 = 0.5 and variance = 10 × 0.05 × 0.95 = 0.475.(c) P (X = 1) = 0.3151 and P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2) = 0.9884.

6.4 (a) p = P (defendant not granted bail) = P (B) = 0.4, n = 20; X = number in sample who arenot granted bail. X has a binomial distribution since there are two outcomes (granted bail ornot), the defendants are independent (random sample), the number of trials is fixed at 20 and theprobability of being granted bail is constant. X ∼ Bin(20, 0.4).(b) P (X = 7) = 0.1659 and P (X > 8) = P (X ≥ 9) = 0.4044.(c) Mean of X = 20 × 0.40 = 8 and variance = 20 × 0.4 × 0.6 = 4.8.

6.5 (a) p = P (victim of an offence) = P (O) = 0.15, n = 20; X = number in sample who have beenthe victim of an offence against their person. X has a binomial distribution since there are twooutcomes (victim or not), the victims are independent (random sample), the number of trials isfixed at 20 and the probability of being a victim of an offence is constant. X ∼ Bin(20, 0.15).(b) P (X = 5) = 0.1028 and P (X ≥ 7) = 0.0219.(b) Mean of X = 20 × 0.15 = 3 and variance = 20 × 0.15 × 0.85 = 2.55.

6.6 (a) p = P (young man already served a sentence) = P (C) = 0.25, n = 12; X = number in samplewho have already served a sentence. X ∼ Bin(12, 0.25).Mean of X = 12 × 0.25 = 3 and variance = 12 × 0.25 × 0.75 = 2.25.(b) P (X = 6) = 0.0401 and P (X > 6) = P (X ≥ 7) = 0.0143.

6.7 (a) p = P (woman in the workforce) = 0.45, n = 20; X = number in sample who are woman.X ∼ Bin(20, 0.45). P (X = 6) = 0.0746.(b) Let Y = number in sample who are men. Y ∼ Bin(20, 0.55). P (Y ≥ 12) = P (X ≤ 8) =1 − 0.5857 = 0.4143.

6.8 p = P (child has a psychological problem) = P (Ps) = 0.10, n = 15; X = number in sample whohave a psychological problem. X ∼ Bin(15, 0.10). P (X = 3) = 0.1285 and P (X > 3) = P (X ≥4) = 0.0556.

6.9 (a) n = 12, X = number without children. P (no children) = P (C) = 0.25. X ∼ Bin(12, 0.25).

66

Two outcomes (children or not), fixed number of trials, independent families (random sample) andp roughly constant.(b) P (X = 5) = 0.1032. P (X < 2) = P (X = 0) + P (X = 1) = 0.1584.(c) Mean number of children = 12× 0.25 = 3. P (X = 6) = 0.04, which is very small, less than 5%and P (X ≥ 6) = 0.0544, which is almost less than 5%. So this outcome (having 6 children) is veryunlikely and suggests that perhaps we did not take a random sample.

6.10 p = P (first admission is due to alcohol abuse) = 0.40, n = 15; X = number in sample where alcoholabuse was the leading cause of their first admission. X ∼ Bin(15, 0.40). P (X < 4) = P (X ≤ 3) =P (X = 0) + · · · + P (X = 3) = 0.0905.

6.11 (a) n = 15 patients. X = number who are extroverted. P (extroverted) = 0.4. Require fixednumber of trials, two outcomes (success and failure), trials must be independent and probabilityof success must remain constant from trial to trial. Then X ∼ Bin(15, 0.4).(b) P (X = 5) = 0.1859. P (X < 5) = 0.2173. P (X > 12) = P (X ≥ 13) = 0.0003.

6.12 n = 8, P (sold by sole agent in first week) = 0.2. X = number sold by sole agent in first week.X ∼ Bin(8, 0.2).(a) P (X = 0) = 0.1678.(b)P (X ≥ 1) = 1 − P (X = 0) = 1 − 0.1678 = 0.8322.(c) P (X = 8) = 0.0000.

6.13 (a) X = number of shells that are plain. n = 20, p = P (plain shell) = 1 − P (streaked shell) = 0.4;X ∼ Bin(20, 0.40). P (X = 9) = 0.1597.(b) P (X ≥ 10) = 0.2447.(c) Proportion of plain shells is 29

40= 0.725. This is well above 0.4 and we would conclude that the

theorised proportions are incorrect. (Or when n = 40, mean = 40 × 0.4 = 16 and 29 is well abovethis and the same conclusion applies.) P (X ≥ 29) = 0.0000 when X ∼ Bin(40, 0.4) so this is avery unlikely result if p = 0.4.

6.14 X = number of registered vehicles that are cars. n = 20, p = P (car) = 0.7; X ∼ Bin(20, 0.7).Since p > 0.5, let Y = number of registered vehicles that are not cars; Y ∼ Bin(20, 0.3).(a) P (X = 20) = P (Y = 0) = 0.0008.(b) You would expect np = 20 × 0.7 = 14 to be cars.(c) P (X = 10) = P (Y = 10) = 0.0308. This is very small, so getting 10 cars out of 20 vehicles isan unlikely event. This would lead us to doubt the randomness of the sampling process.

6.15 (a) p = 4

20= 0.2.

(b) Let X = number of toddlers with score greater than 10.7, then X ∼ Bin(20, 0.2) as there aretwo outcomes (greater than 10.7 or not), n is fixed (n = 20), probability of getting greater than10.7, p = 0.2 is constant and the toddlers are independent.(c) P (X = 14) = 0.0000.(d) P (X ≥ 15) = 0.0000.

6.16 (a) Let X = number of short-life products past their use-by dates, then X ∼ Bin(15, 0.3). Thereare two outcomes (past use-by or not), n is fixed (n = 15), probability (p = 0.3) is constant andthe products are independent (random sample).(b) P (X = 0) = 0.0047.(c) P (X ≥ 2) = 0.9647.

6.17 (a) X ∼ Bin(20, 0.55). X has mean 20 × 0.55 = 11 and standard deviation√

20 × 0.55 × 0.45 =2.225.(b) Since p > 0.55, let Y = number of days out of 20 where there was no increase. Y ∼Bin(20, 0.45); P (X ≥ 10) = P (Y ≤ 10) = 1 − P (Y ≥ 11) = 1 − 0.2493 = 0.7507.(c) Each day’s share price change might not be independent from the previous day’s, there couldbe trends in the market.

6.18 (a) X ∼ Bin(20, 0.1). The mean of X is 20× 0.1 = 2 and variance is 20× 0.1× 0.9 = 1.8, standarddeviation =

√1.8 = 1.342.

67

(b) P (X > 6) = P (X ≥ 7) = 0.0024.

6.19 (a) X ∼ Bin(15, 0.35). P (X = 2) = 0.0476, P (X > 10) = P (X ≥ 11) = 0.0028, P (X < 8) =1 − P (X ≥ 8) = 1 − 0.1132 = 0.8868.(b) Mean = 15 × 0.35 = 5.25, variance = 15 × 0.35 × 0.65 = 3.4125.(c) Mean proportion is p = 0.35 and variance of p = p(1 − p)/n = 0.35 × 0.65/15 = 0.0152.

6.20 (a) The proportion of skua observed was p = 36

45= 0.80.

(b) X ∼ Bin(12, 0.8). Since p > 0.8, let Y ∼ Bin(12, 0.2) where Y = number of quadrats in whichsouthern skua were not observed. X = n − Y . P (X = 6) = P (Y = 6) = 0.0155P (6 ≤ X ≤ 10) = P (X = 6) + · · · + P (X = 10) = P (Y = 6) + · · · + P (Y = 2) = 0.7213.(c) P (Y < 2) = P (Y = 0) + P (Y = 1) = 0.2749. P (Y > 4) = P (Y ≥ 5) = 0.0726.

6.21 (a) Proportion with blood count less than 6 is p = 2

10= 0.2.

(b) X ∼ Bin(20, 0.2). P (X < 4) = P (X = 0) + · · · + P (X = 3) = 0.4114.

6.22 X ∼ Bin(11, 0.4). P (X = 5) = 0.2207 and P (X ≥ 6) = 0.2465.

6.23 (a) X ∼ Bin(12, 0.2). P (X ≥ 3) = 0.4417.(b) P (X = 0) = 0.0687(c) If 7 are not sent to the United States, 5 are sent there, so P (X = 5) = 0.0532.

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Chapter 7

Exercises

The first set of exercises give you important practice using the standard normal table.

7.1 Find the following probabilities:(a) P (0 < Z < 1.3) (b) P (−1.53 < Z < 0) (c) P (−0.8 < Z < 0.5)(d) P (0 < Z ≤ 0.8) (e) P (−1.96 < Z < 1.96) (f) P (−2.5 ≤ Z < 2.5)

7.2 Find the following probabilities:(a) P (0 < Z ≤ 0.95) (b) P (0 ≤ Z < 1.65) (c) P (−0.3 < Z < 0.6)(d) P (−3 < Z < 0) (e) P (−1.2 < Z < 2) (f) P (−1.4 < Z < 1.4)

7.3 Find the following probabilities:(a) P (Z > 1.65) (b) P (Z ≤ −1.2) (c) P (Z < 2)(d) P (Z > −0.9) (e) P (1.2 < Z < 2) (f) P (−3 < Z < −1.5)

7.4 Find z to satisfy:(a) P (Z > z) = 0.025 (b) P (Z < z) = 0.25 (c) P (Z < z) = 0.9(d) P (Z > z) = 0.6 (e) P (−z < Z < z) = 0.8

The following questions give you practice with standardising and the CLT, and more practice using thetables.

7.5 Given X ∼ N(10, 0.25), find the following probabilities:(a) P (X > 10.9) (b) P (9 < X < 11) (c) P (X < 8.7)(d) Find P (X > 10.25) where n = 10.(e) Find x such that P (X > x) = 0.05 where n = 10.

7.6 Given X ∼ N(20.2, 4), find the following probabilities:(a) P (X > 20.9) (b) P (15 < X < 25) (c) P (X < 19)(d) Find P (X < 19) where n = 20. Compare your answer to that in (c).(e) Find x such that P (X < x) = 0.85 where n = 20.

7.7 Lengths of custodial sentences handed out to under 20-year-old males in New Zealand are roughlynormally distributed with mean µ = 4.2 years and standard deviation σ = 2.8 years.

(a) What is the probability that a randomly selected male under 20 given a custodial sentence issentenced to a year or less?

(b) What proportion of the male prison population under 20 have sentence lengths of up to sixyears?

(c) What length of sentence are 80% of 20-year-old men in jail serving less than?(d) Why may the normal distribution not be a very suitable one for these sentences? [Hint:

consider P (X < 0).]

7.8 In families consisting of a couple with children, the age of the mother at the birth of her first child,Y , follows roughly a normal distribution with mean µ = 26.4 and standard deviation σ = 3.2 years.

(a) What proportion of couples with children have had their first child before the mother is 20?(b) By what mother’s age are 75% of first-born children of couples with children born?

7.9 The monthly percentage returns for Telecom New Zealand’s share price is approximately normalwith mean µ = 0.5% and standard deviation σ = 6.3%. In a randomly selected month:

(a) What is the probability of a positive return, i.e. what is P (X > 0)?(b) What is the probability that the return exceeds 4% in a month?

7.10 The distribution of the birth weights of babies delivered at 40 weeks is approximately normal witha mean of 3500g and standard deviation of 430g.

(a) A baby weighing 2500g or less is considered to have low birth weight. What proportion ofbabies delivered at 40 weeks have a low birth weight?

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(b) What is the probability that a randomly chosen mother will give birth to a baby weighingmore than 4000g?

(c) Calculate the upper quartile of birth weight. What is the lower quartile?

7.11 The output voltage from an electronic device has a normal distribution with mean µ = 1.0mV andstandard deviation σ = 0.4mV under the usual operating conditions. If the voltage exceeds 2.0mV,then an alarm sounds.

(a) Calculate the probability that, under usual operating conditions, the alarm rings.(b) Calculate the probability that the voltage is between 0.0mV and 0.5mV.

7.12 The hedge sparrow has a variety of breeding strategies. A sample of 38 hedge sparrows usingmonogamy had an average of five young per female, i.e. x = 5, with s = 2.5. It is claimed that themean number of young per female for monogamous hedge sparrows is µ = 4.2. In samples of sizen = 38, find P (X > 5).(Data based on Davies and Houston, 2003.)

7.13 Acceptable discharge for a particular factory is specified to average 25mg/L of insoluble waste witha standard deviation of 2mg/L. If a random sample of 100 separate one litre specimens is takenfrom the discharge, what is the probability that the sample mean will exceed 25.4mg/L?

7.14 Assume that the average height of 18-year-old male New Zealanders is 1.70m with a standarddeviation of 0.15m. A random sample of 100 18-year-old male New Zealanders is selected, and themean height X is calculated. Calculate P (X > 1.74m).

In the following exercises, we use the normal approximation to the binomial distribution.

7.15 Suppose X ∼ Bin(150, 0.7). Justify the use of the normal approximation, and then calculate thefollowing probabilities:(a) P (X ≥ 100) (b) P (X > 100) (c) P (90 < X < 110)

7.16 Suppose X ∼ Bin(100, 0.4). Justify the use of the normal approximation, and then calculate thefollowing probabilities:(a) P (X ≥ 40) (b) P (X > 32) (c) P (35 < X < 45)

7.17 Seventy-five percent of the members of the Anglican faith in New Zealand are over 20-years-old.In a random sample of 200 New Zealand Anglicans, find the probability that X lies between 140and 180, where X is the number in the sample over 20-years-old.

7.18 Thirty percent of coal miners over 50 years of age have respiratory problems. Estimate the prob-ability that more than 35 miners in a random sample of 80 miners are suffering from respiratoryproblems.

7.19 A psychologist was assisting with the trialling of a new road sign and comparing motorists’ recallof the information on this new sign with that for the existing sign. A new sign was erected and1km past the sign a random sample of n = 80 motorists were stopped and asked to recall theinformation on the sign. Fifty-six of the 80 stopped could correctly recall it.

(a) Can the normal approximation to the binomial be used in this situation? Why?(b) Using p = 0.7 as the actual probability of recalling the information on the sign, find the

probability that in a random sample of 200 motorists more than 75% will correctly recall thenew sign.

The following questions give you practice calculating and interpreting p-values.

7.20 Find the p-values of the following results, which all come from random samples of size n = 81 froma population in which the mean is thought to be 21.5 and the standard deviation is known to be5.3. In each case, after you have worked out the p-value, comment on the implications of this valuefor the claim that µ = 21.5.

(a) Sample mean = 23.1 and it is thought that if µ is not equal to 21.5 then it is likely to be higher.

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(b) Sample mean = 23.1 and it is thought that if µ is not equal to 21.5 there is no indicationwhether it is likely to be higher or lower.

(c) Sample mean = 20.6 and it is thought that if µ is not equal to 21.5 then it is likely to be lower.(d) Sample mean = 23.1 and it is thought that if µ is not equal to 21.5 then it is likely to be lower.

7.21 Find the p-values of the following results, which all come from random samples of size n = 64 froma population in which the proportion having a certain characteristic is thought to be p = 0.4. Ineach case, after you have worked out the p-value, comment on the implications of this value for theclaim that p = 0.4.

(a) The number in the sample with the characteristic is 32 and it is thought that if p is not equalto 0.4 then it is likely to be higher.

(b) The number in the sample with the characteristic is 32 and if p is not equal to 0.4 there is noindication whether it is likely to be higher or lower.

(c) The number in the sample with the characteristic is 17 and it is thought that if p is not equalto 0.4 then it is likely to be lower.

(d) The number in the sample with the characteristic is 32 and it is thought that if p is not equalto 0.4 then it is likely to be lower.

7.22 An insurance company promises that 90% of its claims are settled within 30 days. A consumergroup who believe that the proportion is lower than this takes a random sample of 75 claims andfinds that 55 are settled within 30 days. Find the p-value of their result and decide whether or notit supports the company’s or the consumer group’s view.

7.23 The average hourly wage of 200 18-year-olds employed part-time in various supermarkets in NewZealand is $9.50 with a standard deviation of 50 cents, i.e. $0.5. It has been suggested that part-time workers aged 18 are receiving an average hourly wage of $10.30. Find the p-value of these datafor the 200 supermarket workers. Does their data support the claim about the national part-timeworker wage? Explain your answer.

7.24 Forty percent of Samoans living in New Zealand were not born in New Zealand. Auckland is usuallythe point of entry for Samoans immigrating to New Zealand and because of this it is suggested thatthe proportion of overseas born Samoans will be higher there. In a random sample of 500 Samoansin South Auckland, 175 were born in New Zealand. Find the p-value of this result. Does it supportthe suggestion that the proportion of overseas-born Samoans is higher in South Auckland than inthe country as a whole? Justify your answer.(Census data, Statistics New Zealand, 2001.)

7.25 For many years 16% of males aged between 20 and 24 requiring inpatient psychiatric treatmenthave been admitted with a diagnosis of schizophrenia. In a sample of 40 new admissions to apsychiatric ward in 2003 the number of males admitted with a diagnosis of schizophrenia was nine.Does the p-value of this result indicate that this number is sufficiently greater than 16% for us tochange our view about the proportion of males in this age bracket admitted to a psychiatric wardwho will need treatment for schizophrenia in this year?

7.26 The proportion of people who visit the doctor in any given year is 81%. In a random sample of 78teenage boys, 56 said they had visited their doctor in the last 12 months. Find the p-value of thisresult in order to establish whether this group of people are less likely to visit their doctor thanthe population at large.

7.27 Find the p-value of the result in Example 7.12 when the alternative scenario is just that µ 6= 4.2.Does this indicate that the sample of birds is different from the general population?

Solutions

Note that z-scores are rounded to 2dp to be consistent with using the table in Appendix C.3

7.1 (a) 0.4032; (b) 0.4370; (c) 0.2881 + 0.1915 = 0.4796; (d) 0.2881; (e) 2 × 0.475 = 0.95; (f) 0.9876.

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7.2 (a) 0.3289; (b) 0.4505; (c) 0.1179 + 0.2257 = 0.3436; (d) 0.4987; (e) 0.3849 + 0.4772 = 0.8621; (f)2 × 0.4192 = 0.8384.

7.3 (a) 0.5 − 0.4505 = 0.0495; (b) 0.5 − 0.3849 = 0.1151; (c) 0.5 + 0.4772 = 0.9772; (d) 0.5 + 0.3159 =0.8159; (e) 0.4772− 0.3849 = 0.0923; (f) 0.4987− 0.4332 = 0.0655.

7.4 (a) 1.96; (b) −0.6745; (c) 1.2816; (d) −0.2533; (e) 1.2816.

7.5 (a) P (X > 10.9) = P (Z > 1.8) = 0.5 − 0.4641 = 0.0359.(b) P (9 < X < 11) = P (−2 < Z < 2) = 2 × P (0 < Z < 2) = 2 × 0.4772 = 0.9544.(c) P (X < 8.7) = P (Z < −2.6) = 0.5 − 0.4953 = 0.0047.(d) P (X > 10.25) = P (Z > 1.58) = 0.5 − 0.4429 = 0.0571.(e) P (X > x) = 0.05 implies P (0 < X < x) = 0.45. When p = 0.45, z = 1.6449 and x =10 + 1.6449× (0.5/

√10) = 10.26. So P (X > 10.26) = 0.05.

7.6 (a) P (X > 20.9) = P (Z > 0.35) = 0.5 − 0.1368 = 0.3632.(b) P (15 < X < 25) = P (−2.6 < Z < 2.4) = 0.4953 + 0.4918 = 0.9871.(c) P (X < 19) = P (Z < −0.6) = 0.5 − 0.2257 = 0.2743.(d) P (X < 19) = P (Z < −2.68) = 0.5 − 0.4963 = 0.0037. This probability is much less than thatin part (c) because the sample mean is much closer to µ on average, than a single observation.(e) P (X < x) = 0.85 implies P (0 < X < x) = 0.35. When p = 0.35, z = 1.0364 and x =20.2 + 1.0364× (2/

√20) = 20.66. So P (X < 20.66) = 0.85

7.7 (a) P (X < 1) = P (Z < −1.14) = 0.5 − 0.3729 = 0.1271.(b) P (X < 6) = P (Z < 0.64) = 0.5 + 0.2389 = 0.7389.(c) P (X < x) = 0.80 implies P (0 < X < x) = 0.30. When p = 0.30, z = 0.8416 and x =4.2 + 0.8416× 2.8 = 6.56 years. So P (X < 6.56) = 0.80.(d) P (X < 0) = P (Z < −1.5) = 0.5−0.4332 = 0.0668, so there is a decent probability of a negativesentence when using the normal distribution.

7.8 (a) P (X < 20) = P (Z < −2) = 0.5 − 0.4772 = 0.0228.(b) P (X < x) = 0.75 implies P (0 < X < x) = 0.25. When p = 0.25, z = 0.6745 and x =26.4 + 0.6745× 3.2 = 28.56 years. So P (X < 28.56) = 0.75.

7.9 (a) P (X > 0) = P (Z > −0.08) = 0.5 + 0.0319 = 0.5319.(b) P (X > 4) = P (Z > 0.56) = 0.5 − 0.2123 = 0.2877.

7.10 (a) P (X < 2500) = P (Z < −2.33) = 0.5 − 0.4901 = 0.0099.(b) P (X > 4000) = P (Z > 1.16) = 0.5 − 0.3770 = 0.1230.(c) P (X < x) = 0.75 implies P (0 < X < x) = 0.25. When p = 0.25, z = 0.6745 and UQ =3500 + 0.6745 × 430 = 3790.0g. So P (X < 3790.0) = 0.75. Since the normal distribution issymmetric, LQ = 3500− 0.6745× 430 = 3210.0g.

7.11 (a) P (X > 2) = P (Z > 2.5) = 0.5 − 0.4938 = 0.0062.(b) P (0 < X < 0.5) = P (−2.5 < Z < −1.25) = 0.4938− 0.3944 = 0.0994.

7.12 P (X > 5) = P (Z > 1.97) = 0.5 − 0.4756 = 0.0244.

7.13 P (X > 25.4) = P (Z > 2) = 0.5 − 0.4772 = 0.0228.

7.14 P (X > 1.74) = P (Z > 2.67) = 0.5 − 0.4962 = 0.0038.

7.15 n = 150 ≥ 30; np = 150 × 0.7 = 105 and np ± 3√

np(1 − p) = 105 ± 3 × 5.61 = 88.17, 121.83 andboth lie within 0 and n, so we can use the normal approximation.(a) z = (99.5−105)/5.61 = −0.98, P (X ≥ 100) = P (X ′ > 99.5) = P (Z > −0.98) = 0.5+0.3365 =0.8365.(b) z = (100.5 − 105)/5.61 = −0.80, P (X > 100) = P (X ′ > 100.5) = P (Z > −0.80) = 0.5 +0.2881 = 0.7881.(c) z1 = (90.5 − 105)/5.61 = −2.58 and z2 = (109.5 − 105)/5.61 = 0.80, P (90 < X < 110) =P (90.5 < X ′ < 109.5) = P (−2.58 < Z < 0.8) = 0.4951 + 0.2881 = 0.7832.

7.16 n = 100 ≥ 30; np = 100 × 0.4 = 40 and np ± 3√

np(1 − p) = 40 ± 3 × 4.90 = 25.3, 54.7 and bothlie within 0 and n, so we can use the normal approximation.

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(a) z = (39.5 − 40)/4.90 = −0.10, P (X ≥ 40) = P (X ′ > 39.5) = P (Z > −0.10) = 0.5 + 0.0398 =0.5398.(b) z = (32.5 − 40)/4.90 = −1.53, P (X > 32) = P (X ′ > 32.5) = P (Z > −1.53) = 0.5 + 0.4370 =0.9370.(c) z1 = (35.5 − 40)/4.90 = −0.92 and z2 = (44.5 − 40)/4.90 = 0.92, P (35 < X < 45) = P (35.5 <X ′ < 44.5) = P (−0.92 < Z < 0.92) = 2 × 0.3212 = 0.6424.

7.17 z1 = (140.5 − 150)/6.12 = −1.55 and z2 = (179.5 − 150)/6.12 = 4.82, P (140 < X < 180) =P (140.5 < X ′ < 179.5) = P (−1.55 < Z < 4.82) = 0.4394 + 0.5 = 0.9394.

7.18 z = (35.5 − 24)/4.10 = 2.81, P (X > 35) = P (X ′ > 35.5) = P (Z > 2.81) = 0.5 − 0.4975 = 0.0025.

7.19 (a) n = 80 ≥ 30 and p ± 3√

p(1 − p)/n = 0.7 ± 3 × 0.05 = 0.55, 0.85 and both lie within 0 and 1,so the normal approximation to the binomial applies.(b) z = (0.75 − 0.70)/0.03 = 1.54, P (p > 0.75) = P (Z > 1.54) = 0.5 − 0.4382 = 0.0618.

7.20 The standard error is 5.3/√

81 = 0.5889. For x = 23.1, z = (23.1 − 21.5)/0.5889 = 2.72. Forx = 20.6, z = (20.6 − 21.5)/0.5889 = −1.53.(a) p-value = 0.5 − 0.4967 = 0.0033. This is very small so it is unlikely that µ = 21.5.(b) p-value = 2 × (0.5 − 0.4967) = 0.0066. This is very small so it is unlikely that µ = 21.5.(c) p-value = 0.5 − 0.4370 = 0.0630. This p-value is reasonably small, so we may doubt whetherµ = 21.5 is correct, given our result.(d) p-value = 0.5 + 0.4967 = 0.9967. This large value means that our result is a highly likely oneif µ = 21.5 compared with µ < 21.5.

7.21 The normal distribution is reasonable, since np − 3√

np(1 − p) = 25.3 > 0. For p = 0.5, z =(0.5 − 0.4)/0.061 = 1.63. For p = 17/64 = 0.266, z = (0.266 − 0.4)/0.061 = −2.19.(a) p-value = 0.5 − 0.4484 = 0.0516. Fairly unlikely result to get if 0.4 is the correct proportion.(b) p-value = 2 × (0.5 − 0.4484) = 0.1032. The chance of getting this result if 0.4 is the trueproportion is approximately 10%.(c) p-value = 0.5 − 0.4857 = 0.0143. The chance of getting this result if 0.4 is the true proportionis fairly small so we would suspect that the true proportion is less than 0.4.(d) p-value = 0.5+0.4484 = 0.9484. Highly likely, an entirely reasonable result if p = 0.4 comparedto p < 0.4.

7.22 p = 55/75 = 0.733,√

p(1 − p)/n = 0.035, z = (0.733 − 0.90)/0.035 = −4.811. It is thought that pis lower than 90% and so p-value ≈ 0. This implies that the insurance company’s claim is false, asif p is really 0.9 the data has produced an extremely unlikely result.

7.23 σ/√

n = 0.035, z = (9.5 − 10.3)/0.035 = −22.63, and p-value = 0. This implies the suggestion iswrong, i.e. supermarket employees do not get the national part-time workers’ average hourly wage.

7.24 p = 325/500 = 0.65 not born in New Zealand, compared to p = 0.4. P (Z > 11.41) = 0, indicatinga very unlikely result, so it seems clear that in South Auckland the proportion is much higher than40%.

7.25 p = 9/40 = 0.225, with√

p(1 − p)/n = 0.058, z = (0.225 − 0.16)/0.058 = 1.12 and p-value= 0.5− 0.3686 = 0.1314. The p-value is not less than 5%, which suggests that we cannot reject theproportion could still be 16% even though we had a sample proportion of 22.5%.

7.26 p = 56/78 = 0.718, with√

p(1 − p)/n = 0.044, z = (0.718 − 0.81)/0.044 = −2.07 and p-value= 0.5 − 0.4808 = 0.0192. The p-value is small (< 5%), so it is likely this group do visit the doctorless.

7.27 z = 1.97, p-value = 2 × 0.0244 = 0.0488. This is a small value which means that getting a resultof x = 5 if µ = 4.2 is unlikely, and this suggests that µ is not 4.2.

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Chapter 8

Exercises

The following exercises give practice for large sample inference for a population mean.

8.1 An important health issue especially for poorer countries is the differential in price between brandname pharmaceuticals and their generic equivalents. For example in New Zealand:

Panadol Prozac Zertec

Brand name price ($) 4.95 18.50 28.60Generic version price ($) 2.95 1.93 2.75

A sample of 49 prescription medicines in the brand name price range $15-30 were compared withtheir generic prices. The average difference in price, x, was $24.93 with a standard deviation ofs = $8.74. Construct a 95% confidence interval for µ, the mean price difference for all prescriptionmedicines in this price group.

8.2 Acid rain, due to industrial pollution has been measured in the greater Auckland region. Waterspecimens from 40 different locations during one rainfall gave pH values with a mean x = 5.5381and a standard deviation s = 0.4875. pH is a measure of acidity where 0 = acid, 14 = alkaline andpure rain falling through clean air registers a pH value of 5.7.

(a) Stating any assumptions that are necessary, test the hypothesis H0 : µ = 5.7 against thealternative Ha : µ < 5.7 where µ was the mean pH value for the greater Auckland regionduring that rainfall. Use a 5% significance level.

(b) What is the p-value for your result?

8.3 During a single week in September, 2003, the average number of unjustified absences (one formof truancy) in a sample of 320 schools in New Zealand was x = 118.60 per day with a standarddeviation, s = 79.05.

(a) Use this data to construct a 90% confidence interval for the true mean number of unjustifiedabsences per day in New Zealand schools.

(b) Test the statement that the mean number of unjustified absences per day in New Zealandschools is 125.5 using α = 0.10.

(c) How many schools would have been needed to have been sampled in order to achieve a marginof error of ±4 in your confidence interval?

(d) Suggest some possible factors that could make your confidence interval an unreliable estimateof the true mean number of unjustifiable absences per day in New Zealand schools?

8.4 In an experimental trial, 64 sheep were injected with a particular antibiotic at 10mg/kg bodyweight. After 90 minutes, the mean blood serum concentration of the antibiotic was 24.857µg/mLwith a standard deviation of 3.924µg/mL.

(a) Find a 95% confidence interval for µ, the population mean concentration.(b) Using your result above, determine the sample size required to be 95% confident of a maximum

error of 0.5µg/mL in the estimate of µ.

8.5 In May 2004, monthly sales from a random sample of 38 outlets of a well-known franchise producedx = 3.62 and s = 0.78 (in $million). The previous year in May the mean sales nationally for thisfranchise were $3.94 million.

(a) Is there evidence to suggest that 2004’s mean sales for May for this franchise were less thanfor the previous year at that time? Use α = 1%.

(b) Calculate the p-value for this test and explain its meaning.(c) Give a 99% confidence interval for the true mean sales for this franchise for May 2004.

8.6 The driver of the Subaru in Exercise 2.5 wishes to compare the fuel consumption of his car to thefigure stated by the New Zealand Automobile Association as appropriate for his size car, which is11.26 km/L. Test whether or not his fuel consumption is worse than the recommended value forhis size car, using the sample data given in Exercise 2.5. Use a 5% level for the test.

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8.7 The average number of snapper eggs found in n = 30 trawls by a high-speed surface plankton netover 3000m in moderate sea conditions was x = 674.85 with a standard deviation, s = 359.17.

(a) Construct a 90% confidence interval for µ, the mean number of snapper eggs that would befound in a high-speed surface plankton net in a 3000m trawl in similar conditions.

(b) How many trawls would be required to estimate µ to within 80 eggs with a 90% confidenceinterval?

(c) Does the above data support the assertion that µ = 650 or not? Use a 10% level of significanceand give your reasoning.

8.8 In a study of 50 imprisoned male first offenders aged under 30, the mean age at imprisonmentwas 20.14 years with a standard deviation of 2.42 years. Use this data to test the statement thatnationally the mean age of imprisonment for imprisoned male first offenders under 30, is 19.78.Use a 3% level of significance. State your hypotheses, test statistic, rejection region and conclusioncarefully.

8.9 A random sample of 200 people living in New Zealand with student loans had an average loanbalance outstanding of $14,169 with a standard deviation of $13,352. Use this to find a 95%confidence interval for the true mean value of outstanding student loans. If it is desired to getan estimate of the true value to within ±$500 with 95% confidence, how large a sample would berequired?

This next set of exercises gives practice in intervals and testing for means, based on single small samples.

8.10 In eight samples of water from the Baltic Sea near the port of Gdansk, the following weights ofstrontium (as a % of the total sample weight) were measured: 4.0, 5.5, 4.5, 4.25, 6.0, 5.75, 4.5, 4.0.

(a) Find a 95% confidence interval for the mean percentage by weight of strontium in waters nearGdansk.

(b) Draw a boxplot of this data. What does it suggest about your confidence interval in (a)?

8.11 In past experimentation water samples from the Baltic have had a mean concentration of lead (asa % of the total sample weight) of 18.0. Recent attempts have been made to clean the area of toxicheavy metals in an effort to lower heavy metal pollution levels. In eight samples of water from theBaltic Sea near the port of Gdansk the weight of lead was measured and the following summarystatistics were obtained: x = 17.03, and s = 2.63. Test the claim using a 5% level of significance.What is the approximate p-value of your result?(Data for Exercises 8.10 and 8.11 provided by Ross Renner.)

8.12 The girth in metres of ten kauri trees are: 16.4, 12.4, 10.1, 9.6, 10.6, 9.7, 10.5, 9.7, 9.7, 9.6.

(a) Using this data construct a 90% confidence interval for the mean girth of kauri trees.(b) Approximately what sample size is required to get an accuracy of 0.5m in (a)?(c) In fact, the data above are the girths of the 10 largest kauri. How does this affect your interval

estimate for the mean girth of kauri?

(Data from Lambert and Palenski, “NZ Almanac”, 1983, p263.)

8.13 Over 24 trading days the mean sales volume (in $000/day) for a retail outlet is 6.28 with a standarddeviation of 4.13.

(a) Construct a 99% confidence interval for the mean sales volume per day for this outlet.(b) The data for a subsample of 11 days for the outlet are: 9.41, 2.79, 10.21, 1.08, 7.38, 9.07, 5.03,

6.37, 14.98, 8.66, 5.88. Draw a boxplot of this data. What does the boxplot indicate about thevalidity of the confidence interval in (a)? Justify your answer.

8.14 Lengths (in cm) of a sample of adult cave wetas from the Karori Sanctuary are: 2.1, 2.4, 1.5, 2.2,2.1, 2.6, 1.7, 2.2, 2.6, 1.8, 2.0, 2.4, 1.3, 2.3, 1.9, 1.9, 2.1, 1.8.

(a) Using this sample, test the hypothesis that the mean length of adult cave wetas is 2.15cm. Usea 1% level of significance.

(b) What is the approximate p-value of this test? Why is it approximate?

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8.15 Anxiety levels are usually elevated in persons with low self-esteem, although there are genderdifferences in the extent to which this is true. In a random sample of 24 17-year-old femaleswith known low self-esteem the average anxiety score on a standard test was 7.62 with a standarddeviation of 3.45. Construct a 95% confidence interval for the true mean anxiety score on this scalefor 17-year-old females with self-esteem problems. What is the margin of error associated with thisconfidence interval?

8.16 The driver of the Subaru in Exercise 2.5 wishes to compare the fuel consumption of his car onlong-distance trips to the figure stated by the New Zealand Automobile Association as appropriatefor his size car, which is 11.26 km/L. The long distance fuel consumption figures (km/L) were:11.6, 10.7, 11.5, 10.1, 10.5, 10.9, 11.0, 10.6, 10.6, 10.8, 11.1, 11.6, 11.3, 11.6, 10.5. Test whether ornot his fuel consumption is worse than the recommended value for his size car. Use a 5% level forthe test.

8.17 The average minutes per day spent by non-Maori females in New Zealand on unpaid caring forothers outside the home is 6.0. A random sample of 28 Maori women spent an average of 10.5minutes per day with a standard deviation of 9.2 minutes in unpaid caring for others outside thehome. Is this evidence that Maori women spend more time than non-Maori women in unpaidcaring for others outside the home? Use a 1% level of significance.(Data based on the New Zealand Time Use Survey 1998-1999. Statistics New Zealand)

8.18 The average testosterone level in ng/dL of saliva of 45 to 50-year-old sedentary males is thoughtto be 9.4. A random sample of 15 males in this age group had an average level of 10.2 with avariance of 5.29. Does this information support the accepted level of testosterone in this age groupof males? Test at the 10% level of significance.

The next exercises give practice performing the sign test.

8.19 Twelve prison inmates with no gang affiliations, who were convicted of violent offences, took part inan anger management course. Before the course began the median score on an anger control assess-ment of this group was 42 (out of a maximum of 60). After the six-month course the participants’scores on the same assessment were: 35, 45, 45, 44, 57, 34, 39, 47, 58, 39, 42, 39.

The course is designed to lower anger control scores. Perform a sign test on this suggestion usingα = 0.05. Why is a sign test appropriate for this data?

8.20 Twelve building projects in the Auckland region in 2001 were costed in detail. In 1999 the mediancost of similar type buildings was $0.55 million. The cost (in $million) of these twelve projects in2001 were: 0.45, 0.65, 0.55, 0.49, 0.69, 0.40, 0.39, 0.57, 0.68, 0.49, 0.52, 0.59. The expectation isthat costs have risen. Perform a sign test on this suggestion using α = 0.05.

8.21 Ten participants in a physical fitness programme had their fasting blood sugar levels taken 21days after the programme began. Results were: 92, 108, 104, 102, 93, 85, 90, 93, 88, 103. It hasbeen suggested that the median fasting blood sugar level of people undertaking a physical fitnessprogramme will be lower than the usual median of 105. Perform a sign test to test this suppositionusing α = 0.05 and explain why the sign test is appropriate in this situation.

8.22 One-half of each of seven leaves is treated with one preparation of a virus, the other half withanother. The difference in the number of lesions for the two halves of each leaf is recorded asfollows: 3, 4, 8, −1, 1, 5, 1. Test the hypothesis that there is no difference between the effect of thetreatments using the sign test and a t-test, using α = 0.05. By drawing a boxplot, or otherwise,decide which is the most appropriate of these two tests in this situation.

8.23 Ten small Wellington businesses were surveyed on their profitability in their first year of trading.The profits recorded were (in $000): −17, −8, −4.3, 6.5, 15.3, 18.2, 25, 27.6, 29.9, 87.5. You arehired to test whether or not small Wellington businesses are earning positive profits on average.

(a) Draw a modified boxplot of the data.(b) Use a sign test to test for positive profits. Use a 5% level.

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8.24 The data on the girths in metres of the ten largest kauri trees in Exercise 8.12 are thought to comefrom a distribution that has a median of ten metres. Carry out a sign test on this suggestion withα = 0.1 and a two-sided alternative.

8.25 Fifteen randomly selected house sale prices in the Wellington suburb of Karori during early 2004were (in $000): 235, 280, 289, 290, 310, 315, 335, 340, 340, 365, 480.5, 505, 128, 144, 182. Themedian sales price for the Western suburbs (of which Karori is one) during the same period was$413,000.

(a) Draw a modified boxplot of these sales figures. Is a t-test appropriate for this data?(b) Does the sample indicate that the Karori properties were cheaper than the average? Use the

sign test and α = 5%.

(Thanks to Ron Beccard for the data.)

In this set of exercises, we practice inference for a population proportion based on large samples.

8.26 In the work on memory done by Wade et al (see Exercise 2.22) with a sample of 35 studentsshown five photos, one of which had been doctored to show a hot air balloon, 14 students becameconvinced that they had experienced a hot air balloon ride when young which was not in fact thecase.

(a) Find a point estimate for the proportion of all students who would believe this if shown thephotos.

(b) Is this a situation in which the CLT can be applied? Justify your answer.(c) Using this sample data find a 95% confidence interval for the true proportion of students who

could be induced to have a false memory of an event by using photographs.(d) In her pilot study with just a few people, Wade had found that 50% of her subjects were

able to have a false memory induced. Use your confidence interval to establish whether or notit is likely that 50% of the population can have such a memory planted. Redo this using aformal hypothesis test with α = 0.05, stating hypotheses, test statistic, rejection region andconclusion.

(e) Find the p-value of your test in (d).

8.27 About 1 child in 30 in New Zealand is born with a structural defect. If a sample of 400 childrenborn in Wellington in a year is looked at and the proportion with structural defects is noted, willit be possible to work out a 95% confidence interval for the true proportion with structural defectsin this city, in order for reasonable health services planning to take place? Justify your answer.

8.28 As part of regular monitoring of the national data base for mental health, MHINC (Mental HealthInformation National Collection), 70 mental health files were randomly selected from the records ofa particular District Health Board over a two-month period. Only 40 of these records were foundto have recorded a diagnosis.

(a) Can the CLT be used in this situation? Why?(b) Use this data to find a 95% confidence interval for the national proportion of mental health

records with a recorded diagnosis.(c) One would hope that the true proportion of records that contained a recorded diagnosis would

be at least 80%. Does the data support this? (Test at a 5% level of significance).

(Data from New Zealand Health Intelligence Services)

8.29 Intravenous urography is the X-ray study of kidneys following the injection into a vein of a contrastmedium (CM). The established CM is known to induce severe nausea in 12% of all patients. Anew CM produces much better pictures but an increased nausea rate would be an unacceptableside-effect. In a trial of 70 patients on the new CM, 13 experienced severe nausea.

(a) Find a 90% confidence interval for p, the true proportion of patients in general who wouldexperience severe nausea with the new CM.

(b) Test the hypothesis H0 : p = 12% against the alternative Ha : p > 12% at a 5% level ofsignificance.

(c) What is the p-value of your test in (b)? Explain what your answer means.

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(d) Using the results of this study, how large should the sample be in order to be 90% confidentthat the sample proportion is within 5% (0.05) of the true proportion?

8.30 Of 50 students sampled from Victoria University, 14 exercised regularly.

(a) Calculate a 95% confidence interval for the proportion of all students in that university whoexercise regularly. Why can you be confident of your result?

(b) A similar 95% confidence interval was constructed for the proportion of students at AucklandUniversity who exercised regularly. The resulting confidence interval was (0.29, 0.36). What iswrong with the following statement? “The true proportion of students at Auckland Universitythat exercise regularly is between 0.29 and 0.36 with probability 0.95.”

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8.31 Among 69,224 newborns registered in New Zealand in 2003, 33,937 were girls.

(a) Give a point estimate for p, the proportion of newborns that are girls.(b) Calculate a 95% confidence interval for p, the proportion of newborns that are girls.(c) How large a sample would be needed to reduce the width of the 95% confidence interval to

±0.002?(d) Based on the confidence interval in (b), could you support the statement that a newborn baby

is equally likely to be a boy or girl? Justify your answer.

8.32 In a random sample of 50 possums captured in the Rimutaka Ranges, 23 were found to havetuberculosis.

(a) Can the CLT be used to find a confidence interval for the true proportion of possums carryingtuberculosis in New Zealand? Justify your answer.

(b) Calculate a 99% confidence interval for the proportion of possums in New Zealand carryingthis disease.

(c) What is the margin of error in the confidence interval that you calculated in (b)? How couldthis margin of error be reduced in size?

8.33 A company bases its warranty on the assumption that 80% of customers who buy its microwaveovens will have no repairs within the first two years of purchase. In a sample of 100 of these ovensonly 70 lasted for at least two years without a repair being necessary.

(a) Find a 98% confidence interval for the true proportion of these ovens that will last withoutrepair for two years or more.

(b) How big a sample would you require to obtain the true proportion with a margin of error of±3% at a 98% level of confidence?

8.34 In May 2004, the unemployment rate in New Zealand dropped to a 16-year low of 4.6%. This wasbettered by only four other OECD countries. In Wellington in a random sample of 160 peopleof employable age not involved in tertiary education, seven were found to be unemployed. Doesthis support the widely held view that the unemployment rate in Wellington is lower than in thecountry as a whole? Use α = 0.05.

8.35 A political poll of 500 randomly selected voters gave X = 230 people in favour of the bypassthrough central Wellington.

(a) Find a 90% confidence interval for the true proportion of voters in favour of the bypass. Usingyour confidence interval, is there any suggestion that 50% of voters favour the bypass?

(b) Show, using the results of the poll, that at least 6722 respondents are needed to get theestimated proportion to within ±0.01 of the true value at the 90% level.

8.36 Billboards around Wellington state that 49% of all road accidents in the region happen at inter-sections. In Wellington from December 2003 to May 2004, 145 accidents occurred at intersectionsand 190 did not. Has the rate of accidents at intersections decreased since the introduction of thebillboards in November? Test using a 5% level of significance.(Data supplied by James King, LTSA.)

8.37 There are approximately 530 species of moss in New Zealand of which three are known to be underthreat. Give a point estimate of the proportion of New Zealand mosses that are under threat. Canwe use this to get a 95% confidence interval for the the proportion of New Zealand mosses that areunder threat? If your answer is “yes”, calculate this interval. If your answer is “no”, explain why.

Solutions

8.1 95% CI is 24.93± 1.96 × 8.74/√

49 = 24.93 ± 2.4472 = (22.5, 27.4).

8.2 (a) Test statistic Z = (5.5381 − 5.7)/(0.4875/√

40) = −2.1. Rejection region is values < −1.645.Test statistic is in rejection region, so reject H0.(b) p-value is P (Z < −2.1) = 0.5 − 0.4821 = 0.0179. Needs to have been a random sample oflocations.

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8.3 (a) 118.6 ± 1.6449× 79.05/√

320 = 118.6 ± 7.26886 = (111.3, 125.9).(b) 125.5 is in the 90% confidence interval so we would not reject H0 at the 10% level.(c) 1.6449× 79.05/

√n = 4, so n = (1.6449× 79.05/4)2 = 1056.72, and we would need 1057 schools

to get the required margin of error.(d) September may not be a typical month as the school holidays usually fall within it, absencesmay be more marked in spring and summer. It may be better to take several weeks of data insteadof just one.

8.4 (a) 24.857± 1.96 × 3.924/√

64 = 24.857± 0.961 = (23.9, 25.8).(b) n = (1.96 × 3.924/0.5)2 = 236.608, we would need a sample size of 237 to get the requiredmargin of error.

8.5 (a) H0 : µ = 3.94 vs Ha : µ < 3.94. Test statistic Z = (3.62−3.94)/(0.78/√

38) = −2.53. Rejectionregion is values < −2.33. Z is in rejection region, so reject H0.(b) p-value is P (Z < −2.53) = 0.5 − 0.4943 = 0.0057. There is a very small chance (less than 1%)of getting this sample or one more extreme, if H0 is actually true.(c) 99% CI is 3.62 ± 2.58 × 0.78/

√38 = 3.62 ± 0.326 = (3.29, 3.95).

8.6 H0 : µ = 11.26 vs Ha : µ < 11.26. Test statistic Z = (10.097 − 11.26)/(1.055/√

30) = −6.04.Rejection region is values < −1.6449. Z is in rejection region, so reject H0.

8.7 (a) 90% CI is 674.85± 1.645 × 359.17/√

30 = 674.85± 107.87 = (567.0, 782.7)(b) n = (1.645 × 359.17/80)2 = 54.545, so the sample size would need to be 55 to get a margin oferror of 80 eggs.(c) 650 is in the 90% confidence interval so we would not reject H0 at the 10% level, so µ could be650.

8.8 H0 : µ = 19.78 vs Ha : µ 6= 19.78. Test statistic Z = (20.14 − 19.78)/(2.42/√

50) = 1.052. 3%rejection region is values < −2.1701 or > 2.1701. Z is not in the rejection region, so do not rejectH0 and conclude that the mean age of imprisoned male first offenders under 30 could be 19.87years.

8.9 95% CI is 14169 ± 1.96 × 13352/√

200 = 14169 ± 1850.4927 = (12318.5, 16019.5). n = (1.96 ×13352/500)2 = 2739.46, the sample size would need to be 2740 to get a margin of error of $500.

8.10 (a) x = 4.8125, s = 0.81009 and n = 8 < 30 so 95% CI is 4.8125 ± 2.365 × 0.81009/√

8 =4.8125± 0.6774 = (4.14, 5.49).(b) LQ = 4.125, median = 4.5, and UQ = 5.625 with a highly asymmetric boxplot which castsdoubt on whether the underlying distribution is normal as required by the t-test. The CI is notlikely to be a good estimate.

4.0 4.5 5.0 5.5 6.0

Boxplot of Strontium levels

amount of strontium (%)

8.11 H0 : µ = 18 vs Ha : µ < 18. Test statistic T = (17.03 − 18)/(2.63/√

8) = −1.04. Rejection regionis values < −1.895. T is not in the rejection region so do not reject H0. Using the t7 tables, theapproximate p-value is greater that 10% as the critical value with 10% of the distribution below itis −1.415.

8.12 (a) x = 10.83, s = 2.135 and n = 10 < 30 so 95% CI is 10.83±1.833×2.135/√

10 = 10.83±1.238 =(9.59, 12.07).(b) n = (1.833 × 2.135/0.5)2 = 61.261, so 62 kauri trees would need to be sampled to get anaccuracy of 0.5m.(c) The interval will give an upper bound for the girth of kauri, i.e. the confidence interval values

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will be far too high using only the largest trees.

8.13 (a) n = 24 < 30 so 99% CI is 6.28 ± 2.807 × 4.13/√

24 = 6.28 ± 2.3664 = (3.91, 8.65).(b) LQ = 5.03, median = 7.38, UQ = 9.41. Very symmetric boxplot which supports underlyingnormality. Indicates a valid CI.

2 4 6 8 10 12 14

Boxplot of Sales Volume

sales volume ($000/day)

8.14 (a) x = 2.05, s = 0.354. H0 : µ = 2.15 vs Ha : µ 6= 2.15. Test statistic T = (2.05 −2.15)/(0.354/

√18) = −1.2. α = 1%, so t17 rejection region is values < −2.898 or > 2.898. T

is not in the rejection region so do not reject H0, i.e. the mean length of cave wetas could be 2.15.(b) The approximate p-value is a little greater than 20% as the two-sided 20% t17 values are ±1.333.

8.15 95% CI is 7.62 ± 2.069× 3.45/√

24 = 7.62 ± 1.457 = (6.16, 9.08). The margin of error is 1.457.

8.16 x = 10.96, s = 0.476. H0 : µ = 11.26 vs Ha : µ < 11.26. Test statistic T = (10.96 −11.26)/(0.476/

√15) = −2.439. For α = 5%, t14 rejection region is values < −1.761. T is in

the rejection region so reject H0 in favour of Ha. The evidence supports that the fuel consumptionis worse than the recommended value for this car.

8.17 H0 : µ = 6 vs Ha : µ > 6. Test statistic T = (10.5−6)/(9.2/√

28) = 2.588. α = 1% rejection regionis values > 2.473. T is in the rejection region so reject H0 in favour of Ha. There is evidence thatMaori woman spend more time in unpaid caring for others outside the home.

8.18 H0 : µ = 9.4 vs Ha : µ 6= 9.4. Test statistic T = (10.2−9.4)/(2.3/√

15) = 1.347. t14 rejection regionis values < −1.761 and > 1.761. T is not in rejection region so cannot reject H0. Informationsupports accepted testosterone level.

8.19 H0 : M = 42 vs Ha : M < 42. n = 12. Comparing with 42 and assigning + to over 42:−, +, +, +, +,−,−, +, +,−, 0,−. Discarding the value 42, n = 11 and there are 5 below and 6above m0. Test statistic is X = 6. If H0 true, X ∼ Bin(11, 0.5) and p-value of this result isP (X ≤ 6) = 1−P (X ≥ 7) = 1− 0.2744 = 0.7256 > 0.05 so we do not reject H0. A boxplot of thisdata (with LQ = 39, median = 43, UQ = 46) is asymmetric which casts doubt on any underlyingnormality and thus a sign test is appropriate with this small sample.

8.20 H0 : M = 0.55 vs Ha : M > 0.55. n = 12. Comparing with 0.55 and assigning + to over 0.55:−, +, 0,−, +,−,−, +, +,−,−, +. Discarding the value 0.55, n = 11 and there are 6 below and 5above m0. Test statistic is X = 5. If H0 true, X ∼ Bin(11, 0.5) and p-value of this result isP (X ≥ 5) = 0.7256 > 0.05 so we do not reject H0. If M = 0.55, getting 5 values above and 6values below is a highly likely event.

8.21 H0 : M = 105 vs Ha : M < 105. n = 10. Comparing with 105 and assigning + to over 105:−, +,−,−,−,−,−,−,−,−. There are 9 below and 1 above m0. Test statistic is X = 1. If H0

true, X ∼ Bin(10, 0.5) and p-value of this result is P (X ≤ 1) = 0.0010+0.0098 = 0.0108 < 0.05 sowe must reject H0. A boxplot of this data (with LQ = 90, median = 93, and UQ = 103) is highlyasymmetric, which casts doubt on any underlying normality so you should not use a t-test.

8.22 Sign test: H0 : M = 0 vs Ha : M 6= 0. n = 7. Comparing with 0 and assigning + to over 0:+, +, +,−, +, +, +. There are 1 below and 6 above m0. Test statistic is X = min(1, 6) = 1. IfH0 true, X ∼ Bin(7, 0.5) and p-value of this result is 2 × P (X ≤ 1) = 2 × (0.0078 + 0.0547) =0.125 > 0.05 so we must not reject H0. For the t-test: H0 : µ = 0 vs Ha : µ 6= 0. Test statisticT = (3 − 0)/(3/

√7) = 2.646. t6 rejection region is values < −2.447 and values > 2.447. T is in

the rejection region so we reject H0. There is a difference between the two treatments. LQ = 1,median = 3, UQ = 5, and the boxplot is fairly symmetric, so the t-test is valid in this situation.

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0 2 4 6 8

Boxplot of Difference in Treatment of Leaves

difference in number of lesions

8.23 (a)

−20 0 20 40 60 80

Boxplot of First Year Profitability

profit ($000)

(b) H0 : M = 0 vs Ha : M > 0. n = 10. Comparing with 0 and assigning + to over 0:−,−,−, +, +, +, +, +, +, +. There are 3 below and 7 above m0. Test statistic is X = 7. If H0

true, X ∼ Bin(10, 0.5) and p-value is P (X ≥ 7) = 0.1719 > 0.05 so we must not reject H0.

8.24 H0 : M = 10 vs Ha : M 6= 10. Data gives us 5 above 10, and 5 below. Test statistic is X = 5(number above). If H0 true, X ∼ Bin(10, 0.5). Here the p-value instruction fails, because P (X = 5)is included twice, i.e. we would calculate 2× P (X ≤ 5) = 2× 0.623 = 1.246. In fact, the p-value is1, since if H0 is true, half the observations will be above and half below and this is exactly whatwe observed. We have no evidence to reject H0.

8.25 (a) min = 128, LQ = 235, median = 310, UQ = 340 and max = 505; IQR = 340 − 235 = 105,LQ − 1.5IQR = 235 − 157.5 = 77.5 and UQ + 1.5IQR = 340 + 157.5 = 497.5. 505 is the onlyoutlier and the upper whisker extends out to the next largest observation 480.5

200 300 400 500

Boxplot of House Sale Prices

sale price ($000)

A t-test is probably not appropriate as the data is asymmetric with an outlier.(b) H0 : M = 413 vs Ha : M < 413. n = 15. Comparing with 413, the data gives us 13below and 2 above m0. Test statistic is X = 2. If H0 true, X ∼ Bin(15, 0.5) and p-value isP (X ≤ 2) = 0.0000 + 0.0005 + 0.0032 = 0.0037 < 0.05 so we should reject H0.

8.26 (a) p = 14/35 = 0.4.(b) n ≥ 30 and p± 3

√

p(1 − p)/35 = 0.4± 0.248 which both lie between 0 and 1, so assuming it isa random sample, we can use CLT.(c) 95% CI is 0.4 ± 1.96 ×

√

0.4 × 0.6/35 = 0.4 ± 0.1623 = (0.24, 0.56).(d) 0.5 is inside the CI, supporting H0 : p = 0.5. Test H0 : p = 0.5 vs Ha : p 6= 0.5. Test statisticZ = (0.4 − 0.5)/(

√

0.5 × 0.5/35) = −1.18. Rejection region is values < −1.96 or > 1.96. Teststatistic is not in rejection region, so we cannot reject H0.(e) p-value is 2 × P (Z > 1.18) = 2 × (0.5 − 0.3810) = 0.2380.

8.27 p = 1/30 = 0.033 and n = 400. Need to check if the CLT holds: n ≥ 30, p ± 3√

p(1 − p)/400 =0.033±0.027 which both lie between 0 and 1, so we can use CLT if we assume it is a random sample.

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The 95% CI is 0.0333±1.96×0.009 = 0.0333±0.018 = (0.015, 0.051). The true proportion is likelyto lie within 1.5% and 5.1%.

8.28 (a) p = 40/70 = 0.571 and n = 70. Need to check if the CLT holds: n ≥ 30, p± 3√

p(1 − p)/400 =0.571± 0.177 which both lie between 0 and 1 and it was a random sample so we can use CLT.(b) 95% CI is 0.571 ± 1.96 × 0.059 = 0.571 ± 0.116 = (0.455, 0.687). The true proportion is likelyto lie within 45.5% and 68.7%.(c) 0.80 is not in the 95% confidence interval, so we would reject H0 : p = 0.80 at the 5% level, infavour of p < 0.80. A formal test confirms this.

8.29 (a) p = 13/70 = 0.186. 90% CI is 0.186 ± 1.645√

0.186(1− 0.186)/70 = 0.186 ± 0.0765 =(0.110, 0.263).(b) Test statistic Z = (0.186 − 0.12)/

√

0.12 × 0.88/70 = 1.70. Rejection region is values > 1.645.Test statistic lies in rejection region so we reject H0.(c) p-value is P (Z > 1.70) = 0.5 − 0.4554 = 0.0446. There is a 4.46% chance of getting this resultif H0 is true.(d) Setting 1.645

√

0.186 × (1 − 0.186)/n = 0.05 we get n = 163.9, so required sample size is atleast 164.

8.30 (a) p = 14/50 = 0.28. 95% CI is 0.28 ± 1.96√

0.28(1 − 0.28)/50 = 0.28 ± 0.1245 = (0.156, 0.404).(b) We can’t make a statement about the probability being 0.95. Instead, either “we are 95%confident µ lies in this interval” or “95% of such intervals will contain µ.”

8.31 (a) p = 33937/69224 = 0.490.(b) 0.49 ± 1.96

√

0.49(1− 0.49)/69224 = 0.49 ± 0.004 giving (0.486, 0.494).(c) n = (1.96/0.002)2 × 0.49 × 0.51 = 240003.96, we would need a sample size of 240004.(d) If a newborn baby is equally likely to be a boy or a girl, p = 0.5. 0.5 is above the 95% confidenceinterval, so it is more likely to be a boy than a girl.

8.32 (a) p = 23/50 = 0.46 and n = 50 ≥ 30, p ± 3√

p(1 − p)/50 = 0.46 ± 0.21 which both lie between 0and 1 and it was a random sample, so we can use CLT.(b) 99% CI is 0.46 ± 2.5758

√

0.46(1− 0.46)/50 = 0.46 ± 0.182 = (0.278, 0.642).

(c) The margin of error is 2.5758√

0.46(1 − 0.46)/50 = 0.182, it can be reduced in size if youincrease the sample size.

8.33 (a) 98% CI is 0.70 ± 2.3263√

0.70(1 − 0.70)/100 = 0.70 ± 0.107 = (0.593, 0.807). Note that thisincludes the company’s value p = 0.8.(b) n = (2.3263/0.03)2 × 0.80 × 0.20 = 962.075, we would need a sample size of 963 to get therequired margin of error. Using the sample value p gives a more conservative estimate of n = 1263.

8.34 p = 7/160 = 0.04375 which is too close to zero for normal approximation to be used. The CLTfails as p − 3

√

p(1 − p)/n < 0.

8.35 (a) p = 230/500 = 0.46. 90% CI is 0.46±1.645√

0.46(1 − 0.46)/500 = 0.46±0.037 = (0.423, 0.497).0.5 is above the CI, suggesting that less than 50% of voters favour the bypass.(b) 1.645

√

0.46(1 − 0.46)/6722 = 0.0099 < 0.01 so the estimated value is within 0.01.

8.36 p = 145/(145 + 190) = 0.433. H0 : p = 0.49 vs Ha : p < 0.49. Test statistic Z = (0.433 −0.49)

√

0.49 × 0.51/335 = −2.11. Rejection region is values < −1.645. Test statistic lies in rejectionregion, so we reject H0.

8.37 p = 3/530 = 0.006. n = 530 ≥ 30, p − 3√

p(1 − p)/n = 0.006 −√

0.006(1 − 0.006)/530 =0.006−0.01 = −0.004 < 0, therefore the CLT doesn’t apply, and we can’t find a confidence intervalwithout resorting to the binomial distribution.

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Chapter 9

Exercises

The first set of exercises are based on large sample comparisons of means.

9.1 The playing time of 60 CDs is measured. The 60 CDs were randomly sampled from two populations:30 CDs were selected from a reggae collection and 30 CDs were selected from a hip-hop collection.Summary statistics are:

Reggae CDs n1 = 30 x1 = 54 min s1 = 7 minHip-hop CDs n2 = 30 x2 = 48 min s2 = 8 min

(a) Test the hypothesis that the populations of all reggae and hip-hop CDs have the same meanplaying time using a 1% level of significance. State your hypotheses, test statistic, rejectionregion and conclusion explicitly.

(b) Calculate the p-value of your test.(c) Find a 90% confidence interval for the mean playing time of reggae CDs and a 90% confidence

interval for the mean playing time of hip-hop CDs. Display these on a number line and commenton your graph.

9.2 In a study of the effects of the anxiety level of mothers on child behaviour, the behavioural problemchecklist (BPC) scores were recorded for 48 children whose mothers were classified as “anxious”,and for 53 children whose mothers were regarded as “not anxious”. Summary statistics for theseBPC scores are shown below (high scores represent more behavioural problems than low ones).

Anxious mothers n1 = 48 x1 = 7.92 s1 = 3.45Non-anxious mothers n2 = 53 x2 = 5.80 s2 = 2.87

(a) Test the hypothesis that the children of anxious mothers are more likely to have behaviouralproblems than the children of non-anxious mothers. Use α = 0.05.

(b) Calculate the p-value for this test.(c) Construct a 90% confidence interval for the population mean difference in BPC scores for

children of anxious and non-anxious mothers.(d) If equal sample sizes from both groups of children were taken, what size would this have to be

to get the difference in the population means scores to within 0.7, with 90% confidence?

9.3 The duration of sleep (in hours) of 85 patients suffering from anxiety and 113 normal controls wasmonitored for five successive nights. Summary statistics for the study are:

Normal controls n1 = 113 x1 = 7.82 s1 = 2.19Anxiety patients n2 = 85 x2 = 7.16 s2 = 2.15

(a) Test the hypothesis H0 : µ1 = µ2 against Ha : µ1 > µ2 at the 3% significance level.(b) Determine the p-value of this test.

9.4 Concentration spans (in minutes) of 180 three-year-old children were measured on a particulartask. The 180 children came from two populations: 120 children regularly attended a preschoolcentre (such as kindergarten, creche, playcentre) and 60 did not. The results were as follows:

No preschool n1 = 60 x1 = 11.7 s1 = 4.2Preschool n2 = 120 x2 = 12.5 s2 = 4.6

(a) Test the hypothesis H0 : µ1 = µ2 against Ha : µ1 6= µ2 at the 5% level of significance.(b) Determine the p-value for this test.(c) Find a 95% confidence interval for the difference in concentration spans for children who did

attend preschool and those who did not. Is your result consistent with your test conclusion in(a)? Why is it not very sensible to construct a confidence interval in this situation?

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9.5 A study was conducted to investigate concerns that the construction of a riverside golf course wasaffecting local trout species. Sampling was undertaken downstream of the development and alsoin an unaffected nearby stream. Fish were caught, weighed, tagged and released. The sample datafollows (weights are in kilograms).

Sample mean Standard deviation Sample size

Affected stream 2.67 0.42 35Unaffected stream 2.42 0.45 32

(a) Using α = 3%, test the hypothesis that there is no difference between the two population meantrout weights.

(b) Find the p-value of this test.(c) Form a 95% confidence interval for the difference in the two population mean weights.

9.6 The mean length of stay in hospital after coronary bypass surgery is expected to be the samethroughout the country. Random samples of patients who had undergone coronary bypass surgeryin an Auckland hospital and in a Dunedin hospital are summarised below (stay in days):

Auckland n1 = 48 x1 = 4.2 s21 = 1.01

Dunedin n2 = 34 x2 = 4.8 s22

= 0.53

(a) Test, at a 10% level of significance, whether there is any difference in the population meanlengths of stay in Dunedin and Auckland.

(b) Find a 90% confidence interval for the mean difference in length of stay after coronary bypasssurgery, in Auckland and Dunedin.

9.7 The average corpuscle breadth is assumed to be greater in frogs than in tadpoles. Samples from 66tadpoles gave a sample mean breadth of 15.3 with a standard deviation of 0.86, while in a randomsample of 58 frogs the mean breadth was 15.81 and the standard deviation was 1.7. Does thisdata support the suggestion that the mean breadth of corpuscles for frogs is greater than that fortadpoles? Test at the 1% level of significance.

9.8 A random sample taken in 2003 of 141 men aged between 28 and 33 who were employed, had anaverage wage of $674 per week with a standard deviation of $174.8/week. A sample of 120 suchmen in 2002 had an average wage of $635/week with a standard deviation of $182.1/week. Testwhether the average wage for such men changed significantly between 2002 and 2003 using a 5%level of significance.

In the following problems, means are compared using small samples.

9.9 A study on the effects of a new drug on the duration of seizures in partially controlled epilepticswas undertaken in order to compare it with the effectiveness of a standard anticonvulsant. A smallgroup of volunteers was carefully monitored over 24 hours on the standard anticonvulsant and tenseizures were observed and timed, while another group on the new drug had 13 seizures timed. Itis claimed that the new drug will reduce the duration of seizures. The results were as follows:

Standard treatment New treatment

Mean duration of seizures (min) 4.1 2.6Standard deviation 1.7 0.6Sample size 10 13

(a) Assuming that duration of seizures has a common variability regardless of treatment type, howwould you estimate this common variance?

(b) Test the above claim at a 5% level of significance.(c) Construct a 95% confidence interval for the difference in mean duration times of seizures using

the standard and new anticonvulsants.(d) If the same number of subjects are selected for both treatments (n), how many must there be

to estimate the true difference between population mean duration of seizures to within about±0.05?

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9.10 In Exercise 9.9 the testing was redone with thirteen pairs of epileptic twins, putting one twin ofeach pair on the standard treatment and the other on the new treatment. Seizures were initiatedwith strobe lights and the following data was collected:

Twin family A B C D E F G H I J K L M

Old treatment 3.9 4.1 3.8 4.4 4.6 4.0 3.7 4.2 4.7 3.5 3.2 4.1 4.3New treatment 2.4 2.6 2.4 2.8 2.9 2.5 2.1 2.7 2.9 1.9 1.1 2.5 2.8

Retest the hypothesis that the new treatment reduces seizure duration under these new conditionsusing a matched pairs test at the 5% level of significance. Justify this choice of test.

9.11 Ten water samples were taken from the port of Gdansk and a further eight from Baltiysk. In eachcase the percentage of strontium by weight was measured. The port area is thought to be morehighly polluted by strontium than sea near Baltiysk. The following data was obtained:

Gdansk n1 = 10 x1 = 4.81 s1 = 0.81Baltiysk n2 = 8 x2 = 4.58 s2 = 0.87

(a) In order to establish whether the above claim is correct, what assumptions need to be madeabout the populations that these samples were drawn from?

(b) Assuming that your assumptions in (a) are valid, test the above claim at a 1% level of signifi-cance.

(c) What is the approximate p-value of your result in (b)? Why is your answer an approximation?

(Data provided by Ross Renner.)

9.12 A company is considering whether to close one of its branches and expand one of its two SouthAuckland outlets. The decision has to be made as to which of the two outlets will give a greaterreturn on expansion. Both outlets are thriving, have available space, and both have similar runningcosts. One suggestion has been to expand the one currently most profitable, which entails somerisk as profits could be at a maximum already, but since this is equally true for the other storethis risk is disregarded. Management has speculated that Outlet A would have a lower mean salesvolume as parking is limited in the vicinity. Over 24 trading days the mean sales volume (in $000per day) for each of these two outlets are calculated and the data is as follows:

Outlet A Outlet B

Mean sales volume 6.28 8.58Standard deviation 4.13 5.14Sample size 24 24

(a) Assuming that sales volumes have a common variability regardless of location, how would youestimate this common variance?

(b) The data is to be used to test the above suggestion. State the null and alternative hypothesesand test the null hypothesis at the 5% level of significance.

(c) The data for a subsample of 12 trading days for Outlet A are: 9.41, 2.79, 10.21, 1.08, 7.38,9.07, 5.03, 6.37, 14.98, 8.66, 5.88, 6.06. Draw a modified boxplot of this data.

(d) What assumptions, other than equal variances, are required to ensure that your conclusion in(b) is valid? What does the boxplot indicate about the validity of this remaining assumption?Justify your answer.

9.13 In a study to assess the impact of an industrial development on a nearby river, one of the manyvariables monitored was water temperature. It has been suggested that mean water temperatureis higher in this river than in a similar river 30km away that is not considered to be affected bythe development. Daily temperature readings in degrees Celsius at 3pm were taken for a fortnightin February from each river at the same longitude. Two days’ readings from the “unaffected” riverwere spoiled, but the remaining data are summarised below:

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Unaffected river Affected river

Sample size 12 14Mean temperature 15.4 17.3Standard deviation 0.70 0.73

(a) Assuming that temperature has a common variability in both rivers, form an estimate for thiscommon variance using a pooled variance estimator.

(b) Construct a 95% confidence interval for the difference in mean temperature in the affected andunaffected rivers.

(c) Test the hypothesis that there is no difference in the mean temperatures of these rivers, testingat the 5% level of significance.

(d) If the same number of sample measurements are taken on both rivers (n), how many mustthere be approximately to estimate the true difference between mean river temperatures towithin about ±0.1 degrees with 95% confidence?

9.14 The driver of the Subaru in Exercise 2.5 suspects that his fuel consumption figures are made up oftwo sorts of usage: long-distance travel and around town motoring. The data are:

Long trips

11.6 10.7 11.5 10.1 10.5 10.9 11.0 10.6 10.6 10.811.1 11.6 11.3 11.6 10.5

Around town driving

9.6 8.5 9.7 9.3 8.6 8.2 8.7 9.4 8.5 10.19.6 8.5 9.3 10.3 10.2

(a) Test whether or not the fuel consumption of the car is significantly worse for the around towndriving. Use a 5% level for the test.

(b) Draw boxplots of the two groups. Do the assumptions of your test appear to be met?

9.15 A group of 23 patients with terminal cancer at a particular site on the body, and at a particularclinical stage, are given the standard treatment and have their survival time from diagnosis recordedin months. Another group of 19 such patients are treated with an experimental drug and theirsurvival times are similarly recorded. It is hoped that the new drug will increase survival times.The summary of the data is:

Experimental treatment Standard treatment

Mean survival time 48.11 37.35Standard deviation 14.36 19.09Sample size 19 23

(a) Assuming survival times have the same variability, form a point estimate of this commonvariance.

(b) Test the suggestion above at the 5% level of significance.(c) Construct a 95% confidence interval for the difference in mean survival times for the standard

and experimental treatments.(d) If there had been 19 patients in each group with each patient in the standard group, each

carefully matched with a patient in the experimental group for age, ethnicity, gender andlifestyle, how would you modify your test statistic used in (b)?

9.16 In an experiment to test whether the gender and appearance of an author makes a person feel moreor less convinced of the validity and usefulness of an article, a group of 18 subjects were shown anarticle giving health advice with a photo of the supposed author attached. This photo was of ahealthy looking male in his thirties. Another group of 18 subjects were shown the same article butthis time the photo was of a female in her fifties who was very thin. The participants rated thearticle for usefulness and believability and an overall credibility score was obtained for the articleas shown (the higher the score the more credible the article):

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Male “author” Female “author”

Mean credibility score 9.11 6.55Standard deviation 1.997 1.65Sample size 18 18

(a) What assumptions need to be made about the data before a test of this suggestion can bemade?

(b) Test the above suggestion at the 10% level of significance.(c) Construct an approximate 90% confidence interval for the difference in mean credibility scores

for the article when apparently written by this male and female author. Why is this confidenceinterval only an approximation?

(d) What other test could have been used in this situation?

9.17 New Zealand native frogs live a long time (up to 30 years), and continue to grow throughout theirlives. In 1984/5, 100 frogs (Leiopelma pakeka) were translocated from a patch of bush on MaudIsland to a new habitat at Boat Bay. The following data set summarises the weight gain (in grams,over the 12 years) of frogs found and weighed both in 1984/5 and 1997. Forty-four were still in theoriginal location, and seven were in Boat Bay.

Old habitat n1 = 44 x1 = 1.901 s1 = 1.468New habitat n2 = 7 x2 = 3.95 s2 = 2.161

Perform a t-test at the 5% level of significance to see if the new habitat is better than the old.(Data from Bell, Ben D., Shirley Pledger and Paulette L. Dewhurst (2004). The fate of a populationof the endemic frog Leiopelma pakeka (Anura: Leiopelmatidae) translocated to restored habitat onMaud Island, New Zealand. New Zealand Journal of Zoology 31: 123–131.)

9.18 Possums in the Orongorongo Valley near Wellington have been live trapped, tagged and tattooed.The following is a sample trapped in March 2004, with their weight recorded (in grams). Allpossums in this data set have year of birth 2001.

Females Males

2200 2250 1900 1800 2000 1925 1700 1850 1750 23001925 1650 2000 2200 1400 2650 2550 1850 1800 22752650 1950 1800 1000 1750 23502300

(a) Assuming common variability in weight for males and females, estimate this common variancewith a pooled variance.

(b) Perform a t-test on the data to establish whether or not there is a difference in mean weightfor male and female possums of the same age. Use α = 0.05.

(c) Redo the test in (b) using a Mann-Whitney U -test.(d) Which of the tests above is likely to be more reliable in this situation? Why?

(Data thanks to Murray Efford of Landcare Research.)

9.19 In a study on the cognitive difficulties associated with Parkinson’s Disease, 12 Parkinson’s sufferersand 12 non-Parkinson’s controls of the same age, all take the Mini Mental Status Examination(MMSE), a test of basic mental functioning that is used as a screen for dementia. The summaryof scores on this test are as follows:

Control group Parkinson’s sufferers

Mean x 28.58 27.42Standard deviation, s 2.065 1.782n 12 12

Test whether or not Parkinson’s disease patients score lower on the MMSE than non-Parkinson’scontrols of the same age. Use a 10% level of significance.(Data provided by Jared Smith.)

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The next set of exercises involve F -tests of equality of variance for two populations.

9.20 In Exercise 9.11, the problem relied on the populations of water samples from both locations havingthe same variance. Use the data from that exercise to test this assumption at the 10% level ofsignificance.

9.21 In Exercise 9.9, test your assumption of equal variances at the 5% level of significance.

9.22 In Exercise 9.13, the hypothesis test required an assumption of common variability for the conclu-sion to be valid. Formally test this hypothesis of common variability at the 5% level of significance.

9.23 Test the assumption of equal variances for the fuel consumption data in Exercise 9.14.

9.24 In Exercise 9.18 equal variances in possum weights for the two sexes was presumed. Test whetherthis is the case at the 5% level of significance against the alternative that female possums havea greater variance. Does this alter your confidence in the result of Exercise 9.18? Justify youranswer.

This set of exercises are primarily for the Mann-Whitney U -test.

9.25 As part of a battery of psychological tests, two tests of spatial ability were given to two independentsamples of seven subjects each, and the scores were as follows:

SubjectA B C D E F G

Score on Test I 101 89 112 105 90 91 89Score on Test II 113 89 121 99 104 94 99

The two tests are supposed to measure the same ability.

(a) Give reasons why a two-sample t-test may not be appropriate to test the above claim.(b) Perform a Mann-Whitney U -test on the data at the 5% level of significance.(c) If the values 113 and 121 of the scores on Test II were recorded wrongly and should have been

118 and 127, how would this alter the conclusion of the Mann-Whitney U -test? Give a reasonfor your answer (no calculation necessary).

(d) Actually, the two samples contained the same individuals. Retest the hypotheses in light ofthis new information.

9.26 A group of patients with a certain disease is divided randomly into two groups. Group 1 is treatedwith drug A for one month and group 2 is treated with drug B for the same period. The bloodpressure of each patient is recorded twice daily and the total number of times it exceeds a specified“safe” level in the month is recorded in the following table:

Group 1 13 7 11 20 12 8 9 13Group 2 7 3 10 6 6 8 4 12 11 5

(a) It has been suggested that the two drugs have an equivalent effect when it comes to reducingblood pressure. This hypothesis is to be tested at the 5% level of significance. Perform aMann-Whitney U -test on the data.

(b) If the value 20 in group 1 should really have been 30, would this alter the conclusion of thetest?

9.27 In order to compare the average yields of two varieties of bulgar wheat, nine test plots of varietyA and ten of variety B are compared for yield.

Variety A: 92.2 90.3 85.7 85.2 89.1 87.4 91.5 85.6 91.6Variety B: 82.9 96.1 94.0 87.7 102.0 87.8 100.2 86.0 88.5 89.8

Use a Mann-Whitney U -test to make this comparison with α = 0.05.

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9.28 Fourteen towns of approximately equal size are chosen for a traffic safety study. Eight of thesetowns are randomly selected and in these towns a series of articles on traffic safety are run in thelocal newspaper for a six-month period. The number of reported accidents in the week followingthe safety campaign is:

Safety articles group 4 7 10 11 17 18 19 21Control group 6 9 12 13 13 16

Use a Mann-Whitney U -test to decide whether this campaign has been effective at the 5% level ofsignificance.

9.29 A study was undertaken on the number of victims of violent crime in two police areas known tobe comparable in terms of social indicators: Lower Hutt and Waitakere. The following data wascollected over a six-week period:

Week1 2 3 4 5 6

Lower Hutt 4 7 9 12 13 13Waitakere 10 13 14 14 16 21

Perform a Mann-Whitney U -test on whether the median numbers of victims of violent crime arethe same in the two areas using α = 0.05.

9.30 The actual data from Exercise 9.17 is as follows:

Original location

1.67 0.25 0.43 2.33 2.85 2.22 1.07 0.10 0.85 2.200.50 2.72 1.65 4.20 1.27 2.93 0.80 0.58 -0.30 3.68-1.70 1.66 3.50 0.00 4.10 1.10 3.07 3.15 1.20 1.803.25 1.50 1.97 1.18 1.27 1.48 2.20 5.97 4.18 2.932.40 2.55 3.15 -0.27

New Boat Bay location

2.55 3.80 2.30 4.45 8.50 3.60 2.45

Using the normal approximation, perform a Mann-Whitney U -test to establish whether the newlocation is any different for the frogs (in terms of median weight gain) than the old at a 1% levelof significance. Why might the normal approximation not be very good in this instance?

9.31 A random sample of sales figures for three-bedroom residential apartments in each of CentralWellington and Central Auckland from March-May 2004 are (in $000):

Wellington 1700 460 655 600 570 493 475 290 370 385

Auckland 670.5 498 255 92 100 1700 310 360 360 490

(a) Draw modified boxplots of each of these two samples.(b) Test for a difference in medians using the Mann-Whitney U -test, at the 5% level.

(Data thanks to Ron Beccard.)

9.32 The data on skinks first met in Exercise 3.18 is partially as follows:

Pasture

4 3 0 2 2 1 4 1 2 5 01 5 6 5 6 11 3 1 1 4 85 14 6 8 10 7 4 8 13 6

Replanted forest

15 24 31 8 4 18 14 33 11 16 201 17 12 27 26 18 6 12 16 11 813 12 11 8 10 17 29 3 12 5

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Perform a Mann-Whitney U -test to establish whether the median number of skinks per trap is thesame in the two habitats: pasture and replanted forest. Use a 1% level of significance. Repeat thetest using the normal approximation.

The following data sets involve dependent samples and give practice for paired comparisons.

9.33 A study of red beech (Nothofagus fusca) was undertaken in the Nelson Lakes area. Six locationsall at the same altitude were randomly chosen in the St Arnaud Range. At each location a quadratwas laid down, and then a second, 200m higher than the first. The number of red beech trees(including saplings and seedlings) were counted in each quadrat giving the following data:

Lower altitude 17 19 22 21 28 19Higher altitude 8 14 18 17 17 12

It is expected that the mean number of red beech trees per quadrat will decline with altitude.Using a matched pairs test with α = 0.05, test this supposition. Why is a matched pairs testsensible in this situation?

9.34 In Tongariro National Park, New Zealand mistletoe is under threat from possums. Ten quadratswere placed at random at a fixed altitude. Each quadrat was left in place for a year, with halfof each quadrat given possum protection (in the form of fencing). At the end of this period themistletoe in the quadrats was scored for the amount of leaf browse from 1 = almost none to 5 =almost total and resulted in the following data:

Unprotected half 3 3 5 4 1 3 5 2 5 4Protected half 2 1 3 2 1 5 3 1 2 3

Test whether or not the protection is effective in terms of reducing leaf browse at a 10% level ofsignificance using a matched pairs test.

9.35 It was widely assumed that visitor numbers to New Zealand would be affected by the September11, 2001 terrorist attacks. The following data, looks at five months of visitor arrivals before andafter this date.

January February March April May

Arrivals 2001 6380 7135 5706 5106 3579Arrivals 2002 6604 7580 6532 4796 3813

Test the above assumption at the 5% level of significance, choosing an appropriate test statistic.(Data thanks to David Lowrie, Statistics New Zealand.)

9.36 The metabolic rate of animals is often measured by oxygen consumption in a closed chamber.Resting metabolic rate was measured in the Marlborough green gecko (Naultinus manukanus) overfive trials on successive days, to find if the animals needed conditioning to accustom them to thechamber before their resting metabolic rate could be reliably measured. The following partial dataset gives the metabolic rate readings at which each individual settled enough for a reading to betaken, at the first and second trials. Use a matched pairs test to establish whether the data presentevidence that the animals settled to a lower rate of oxygen consumption at the second trial? Themeasurement is volume (mL) of oxygen per hour per unit body mass (g). Vol1 represents the firsttrial, Vol2 the second.

Gecko 1 2 3 4 5 6 7 8 9 10

Vol1 0.150 0.061 0.126 0.098 0.109 0.076 0.130 0.062 0.075 0.099Vol2 0.132 0.058 0.105 0.078 0.080 0.085 0.103 0.068 0.074 0.080

Gecko 11 12 13 14 15 16 17 18 19 20

Vol1 0.073 0.147 0.150 0.063 0.086 0.079 0.192 0.084 0.102 0.090Vol2 0.076 0.067 0.149 0.060 0.080 0.078 0.140 0.064 0.107 0.097

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(a) Why is a matched pairs an appropriate test in this situation?(b) Perform the matched pairs test at the 5% level of significance stating carefully your hypotheses,

test statistic and conclusion.(c) Redo using a sign test on the differences in metabolic rate from trial 1 to trial 2 using a 5%

level of significance. What do you conclude?

(Thanks to Kelly Hare and Shirley Pledger for this data.)

9.37 A test rating a person’s sense of humour on a scale from 0 to 100 was given to 15 married couples,the scores obtained being:

Wives 49, 88, 51, 50, 86, 41, 52, 69, 83, 89, 77, 87, 65, 44, 92Husbands 56, 90, 38, 47, 85, 40, 55, 58, 68, 74, 83, 62, 60, 31, 89

Use the sign test to decide, on the basis of the data, whether there is a difference in the averagesense of humour of husbands and wives.

9.38 Referring again to the experiment conducted by Wade et al described in Exercise 2.22, in questionD the subjects had to respond to “I believe the event in my memory really occurred in the wayI remember it and that I have not imagined or fabricated anything that did not occur,” and inquestion F to: “As I remember the event, I am extremely confident that the event occurred.” Inboth cases a low score corresponds to a low belief or confidence that the event really occurred. Thescores (out of 7) of ten of the subjects on these two questions were as follows:

Question D 5 4 5 5 6 7 6 3 2 6Question F 3 6 5 6 6 7 4 4 5 6

(a) Perform a sign test on whether the median scores on these two questions are the same usingα = 0.05.

(b) Why is a sign test appropriate here and not a t-test?

(Thanks to Kimberley Wade for the data.)

9.39 The following data are the percentage returns on the top ten New Zealand stocks (NZSX10), andalso for the Australian Stock Exchange (ASX) for 1990-2004.

Year 1990 1991 1992 1993 1994 1995 1996 1997

NZSX10 −0.06 −0.25 0.02 0.04 0.26 0.03 0.02 0.14ASX 0.00 0.00 0.08 0.05 0.16 −0.01 0.13 0.18

Year 1998 1999 2000 2001 2002 2003 2004

NZSX10 −0.22 0.00 −0.02 −0.07 0.07 0.03 0.10ASX −0.04 0.15 0.04 0.07 −0.05 −0.05 0.15

(a) Conduct a matched pairs test to see whether investors in the New Zealand stocks got signifi-cantly lower returns than investors in the Australian stocks. Use α = 5%.

(b) Test the hypotheses in (a) again, using a sign test at the 5% level.(c) Draw side-by-side boxplots of the data, and use these to determine which of your tests was

most appropriate.

The final set of exercises give practice of inference for proportions.

9.40 Longer post-operative recovery times demand more nursing care and drive up health-care costs. Astudy involving 327 patients in a large hospital showed that 135 out of 218 non-smokers and 84out of 109 smokers spent more than one hour in a recovery ward after surgery.

(a) Test the hypothesis that, in general, the proportion of smokers requiring more than one hourin a recovery ward is greater than the corresponding proportion of non-smokers. State the nulland alternative hypothesis clearly, and use a significance level of 1%.

(b) Find the observed significance level (p-value) of your result.

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(c) Calculate a 95% confidence interval for the difference in the proportions of smokers and non-smokers requiring more than one hour in recovery ward after surgery.

9.41 In 1992, 273 of the 645 road deaths in New Zealand involved drink driving as a factor. In 2002, 109of the 404 deaths involved drink driving. Does this indicate a significant decrease in the proportionof fatalities that involve drink driving? Test at the 5% level.(Data source: Land Transport Safety Authority.)

9.42 In a study to assess the effectiveness of using cuddly animals in television advertising, each of100 subjects was shown a series of new television commercials for several consumer products.The subjects were randomly allocated into two groups of 50 people each. A commercial for oneparticular product featured a group of three small kittens and was shown to Group A. A secondcommercial for the same product, identical in every respect to the first, except that the kittenshad been replaced by pigs, was shown to Group B. At the end of the experiment, all subjects wereasked to recall the products that were being advertised from a description of each commercial. InGroup A, 37 subjects correctly recalled the product while in Group B, 23 subjects did. Do theseresults indicate that a group of kittens is more likely to fix a product in a viewer’s memory thanan equivalent group of pigs? Use a 5% significance level and list any assumptions.

9.43 The number of state houses in a sample of 212 houses in New Brighton, Christchurch was ten,while in a sample of 270 households in Tauranga, six were state houses. Is this sufficient evidenceto establish whether the proportion of state houses in New Brighton is the same as the proportionof state housing in Tauranga? Test at α = 0.10.

9.44 In a random sample of 500 Australians, 100 were under the age of 15 while in a random sampleof 200 New Zealanders, 44 were aged below 15. Has New Zealand a younger population thanAustralia?

(a) Calculate p∗, the pooled sample proportion.(b) Use p∗ to answer the question above, testing at a level of significance of 5%.

(Data from New Zealand in Profile, 2004. Statistics New Zealand)

9.45 Of violent offenders in New Zealand prisons in 1999, the proportion of females in prison for homicide(30.4%) was higher than the proportion of men in prison for homicide (21.6%). There were 79women in jail for violent offences of which 24 were for homicide, while the corresponding figuresfor men were 1817 and 392.

(a) By treating 1999 as if it were a sample from many years of prison inmate data, use this datato test whether there is any significant difference in the proportions of men and women in jailfor violent offences who are imprisoned for homicide. Test at the 10% level of significance.

(b) Does this data mean that you should be more afraid of meeting a woman in a dark alley thana man? Explain your answer.

(Data from Census of Prison Inmates 1999. Department of Corrections, 2000)

9.46 Overall, Maori suffer more than non-Maori in New Zealand when it comes to being victims ofcrime. In the New Zealand National Survey of Crime Victims 2001, (Ministry of Justice, 2003) asample of 3568 Europeans and 947 Maori was taken (along with proportionate sized samples fromother New Zealand ethnic groups). The percentage of Maori that had been burgled in 2000 was6.6%, while for Europeans it was 5.2%. By first converting the percentages into the appropriatenumber (round to whole numbers) and finding p∗, the pooled sample proportion, use this data totest whether these differences are significant and whether burglary follows the overall pattern. Usea 1% level of significance.

9.47 In 2004, 52 cyclists in Northland were observed wearing helmets, while an additional 9 were observedwithout helmets. A similar survey conducted on the West Coast of the South Island identified 115out of 121 observed cyclists wearing helmets.

(a) Test, using a 5% significance level, for a difference in the proportion of cyclists wearing helmetsin these two regions.

(b) Check the assumptions of this test, and comment on the validity of the test.

93

(Data source: Land Transport Safety Authority.)

9.48 The Dominion Post reported on 25 March 2004 that the proportion of babies delivered by caesariansection at Wanganui District Health Board had increased from 18.6% in 2002 to 19.9% in 2003.The actual numbers of caesarian births were 124 out of 667 in 2002 and 135 from 678 in 2003. Wasthe increase in the sample proportions significant?

Solutions

9.1 (a) H0 : µ1 = µ2 vs Ha : µ1 6= µ2. Test statistic is Z = (54 − 48)/√

72/30 + 82/30 = 3.0915.Rejection region is values < −2.5758 and > 2.5758, so reject H0, i.e. the mean playing time forreggae and hip-hop CDs differ.(b) p-value = 2 × P (Z > 3.09) = 2 × (0.5 − 0.499) = 2 × 0.001 = 0.002.(c) 90% CI for µ1 is 54±1.6449×7/

√30 = 54±2.102 = (51.9, 56.1). 90% CI for µ2 is 48±1.6449×

8/√

30 = 48 ± 2.403 = (45.6, 50.4). These two confidence intervals do not overlap which supportsthe conclusion in (a), that the mean playing times differ.

Confidence intervals for the individual population means

mu

Reggae CDs

Hip−hop CDs

44 46 48 50 52 54 56 58

9.2 (a) H0 : µ1 = µ2 vs Ha : µ1 > µ2 where 1 = anxious.Test statistic is Z = (7.92−5.80)/

√

3.452/48 + 2.872/53 = 3.34. Rejection region is values > 1.645,so reject H0.(b) p-value = P (Z > 3.34) = 0.5 − 0.4996 = 0.0004.(c) 90% CI for µ1 − µ2 is 7.92 − 5.80 ± 1.645× 0.635 = 2.12 ± 1.045 giving (1.075, 3.165).(d) Require 1.645

√

3.452/n + 2.872/n = 0.7 giving n = 111.22, so need at least n = 112.

9.3 (a) H0 : µ1 = µ2 vs Ha : µ1 > µ2. Test statistic is Z = (7.82 − 7.16)/√

2.192/113 + 2.152/85 =0.66/0.311 = 2.121. Rejection region is values > 1.8808, so reject H0 in favour of Ha, normalpatients get more sleep than those suffering from anxiety.(b) p-value = P (Z > 2.12) = 0.5 − 0.4830 = 0.017.

9.4 (a) H0 : µ1 = µ2 vs Ha : µ1 6= µ2. Test statistic is Z = (11.7 − 12.5)/√

4.22/60 + 4.62/120 =−0.8/0.6858 = −1.1665. Rejection region is values < −1.96 and > 1.96, so we do not reject H0 atthe 5% level, i.e. there is no evidence of a difference in concentration span between the two groups.(b) p-value = 2 × P (Z > 1.17) = 2 × (0.5 − 0.3790) = 2 × 0.121 = 0.242.(c) 95% CI for µ1 − µ2 is −0.8 ± 1.96 × 0.6858 = −0.8 ± 1.344 giving (−2.144, 0.544). This isconsistent with (a) as 0 is in the confidence interval, so it is possible that there is no difference inmean concentration spans. It is not sensible to construct a CI because we have accepted in (a)that there is no difference in means.

9.5 (a) H0 : µ1 = µ2 vs Ha : µ1 6= µ2. Test statistic is Z = (2.67 − 2.42)/√

0.422/35 + 0.452/32 =0.25/0.1066 = 2.34. Rejection region is values < −2.1701 and > 2.1701, so we reject H0 at the 3%level, i.e. there is a difference between the two mean trout weights.(b) p-value = 2 × P (Z > 2.34) = 2 × (0.5 − 0.4904) = 2 × 0.0096 = 0.0192.(c) 95% CI for µ1 − µ2 is 0.25 ± 1.96 × 0.1066 = 0.25 ± 0.2089 = (0.041, 0.459).

9.6 (a) H0 : µ1 = µ2 vs Ha : µ1 6= µ2. Test statistic is Z = (4.2 − 4.8)/√

1.01/48 + 0.53/34 =−0.6/0.191 = −3.135. Rejection region is values < −1.6449 and > 1.6449, so we reject H0 at the10% level.(b) 90% CI for µ1 − µ2 is −0.6 ± 1.6449× 0.191 = −0.6 ± 0.314 = (−0.914,−0.286).

94

9.7 H0 : µt = µf vs Ha : µt < µf . Test statistic is Z = (15.3− 15.81)/√

0.862/66 + 1.72/58 = −2.065.Rejection region is values < −2.326, so we cannot reject H0.

9.8 H0 : µ1 = µ2 vs Ha : µ1 6= µ2. Test statistic is Z = (674 − 635)/√

174.82/141 + 182.12/120 =39/22.20 = 1.756. Rejection region is values < −1.96 and > 1.96, so we do not reject H0 at the5% level; there was no difference in mean wage in 2002 and 2003.

9.9 (a) s2p = (9 × 1.72 + 12 × 0.62)/(10 + 13 − 2) = 1.444.

(b) H0 : µ1 = µ2 vs Ha : µ1 > µ2 where 1 = standard treatment. Test statistic is T = (4.1 −2.6)/

√

1.444/10 + 1.444/13 = 2.967. The t21 rejection region is values > 1.721, so reject H0.(c) 95% CI for µ1 − µ2 is 1.5 ± 2.08 × 0.5054 = 1.5 ± 1.05 giving (0.45, 2.55).(d) Assuming n will be greater than 30, we use the z-score 1.96 for the 95% CI, so we require1.96

√

1.444/n + 1.444/n = 0.05 giving n = 4437.8, so need at least n = 4438.

9.10 The differences between old and new treatments for each twin family: 1.5, 1.5, 1.4, 1.6, 1.7, 1.5,1.6, 1.5, 1.8, 1.6, 2.1, 1.6, 1.5. xd = 1.6077 and sd = 0.1801. H0 : µd = 0 vs Ha : µd > 0. Teststatistic is T = 1.6077/(0.1801/

√13) = 32.1856. The t12 rejection region is values > 1.782, so

reject H0, i.e. the new treatment does decrease the duration of the seizures.

9.11 (a) Since n1 < 30 and n2 < 30, we need to assume underlying normality for both samples and acommon variance.(b) s2

p = (9×0.812+7×0.872)/(10+8−2) = 0.7002. H0 : µ1 = µ2 vs Ha : µ1 > µ2. Test statistic is

T = (4.81− 4.58)/√

0.7002/10 + 0.7002/8 = 0.23/0.397 = 0.579. The t16 rejection region is values> 2.583, so we can’t reject H0 at 1% level, i.e. there is no evidence of a difference in pollution levelsin the two areas.(c) p-value is greater than 10%, as the 10% value (1-sided) for t16 is 1.337 and 0.579 is to the leftof this value.

9.12 (a) s2p = (23 × 4.132 + 23 × 5.142)/46 = 21.73825.

(b) H0 : µA = µB vs Ha : µA < µB. Test statistic is T = (6.28− 8.58)/√

21.738/24 + 21.738/24 =−2.3/1.346 = −1.71. The t46 ≈ t40 rejection region is values < −1.684, so we reject H0 at 5%level.(c) min = 1.08, LQ = 5.455, median = 6.875, UQ = 9.24 and max = 14.98. IQR = 9.24− 5.455 =3.785, LQ − 1.5IQR = 5.455− 5.6775 = −0.2225 and UQ + 1.5IQR = 9.24 + 5.6775 = 14.9175 so14.98 is the only outlier and the upper whisker extends out to 10.21.

2 4 6 8 10 12 14

Boxplot of Outlet A Sales Volume

sales volume ($000)

(d) Need to assume underlying normality, which is checked by looking for symmetry in the boxplot.The boxplot appears quite asymmetric so we may doubt the validity of the result in (b). Also, thesample has an outlier.

9.13 (a) s2p = (11 × 0.702 + 13 × 0.732)/(12 + 14 − 2) = 0.5132.

(b) 95% CI for µ1 − µ2 with t24 = 2.064 is (15.4 − 17.3) ± 2.064 ×√

0.5132/12 + 0.5132/14 =−1.9 ± 0.582 = (−2.482,−1.318).(c) H0 : µ1 = µ2 vs Ha : µ1 6= µ2 where 1 = unaffected river. 0 is not in the 95% confidence intervalin (b), so we would reject H0, and conclude that the mean temperature of the rivers is different.(d) If we assume n will be more than 30, we can use z = 1.96 for the 95% confidence interval, son = (2× 1.962 × 0.5132)/0.12 = 394.3081, and the sample size would need to be at least 395. Thisclearly exceeds 30, justifying our use of z = 1.96.

9.14 (a) For sample 1 = long trips: x1 = 10.96, s1 = 0.476, n1 = 15. Sample 2 = around town driving:

95

x2 = 9.233, s2 = 0.693, n2 = 15. s2p = (14×0.4762+14×0.6932)/(15+15−2) = 0.353. H0 : µ1 = µ2

vs Ha : µ1 > µ2. Test statistic is T = (10.96−9.233)/√

0.353/15 + 0.353/15 = 1.727/0.217 = 7.96.The t28 rejection region is values > 1.701, so we reject H0 at 5% level.(b) Both samples appear to be skewed slightly, The around town range = 2.1, while long trip range= 1.5, these are slightly different, so we may doubt our conclusion in (a).

Long

trip

sA

roun

d to

wn

8.5 9.0 9.5 10.0 10.5 11.0 11.5

Side by Side Boxplots for Fuel Consumption

fuel consumption (km/L)

9.15 (a) s2p = (18 × 14.362 + 22 × 19.092)/(19 + 23 − 2) = 293.23.

(b) H0 : µ1 = µ2 vs Ha : µ1 > µ2 where 1 = experimental treatment. Test statistic is T =(48.11 − 37.35)/

√

293.23/19 + 293.23/23 = 2.03. The t40 rejection region is values > 1.684, soreject H0.(c) 95% CI for µ1 − µ2 is 10.76 ± 2.021 × 5.3086 = 10.76± 10.73 giving (0.03, 21.49).(d) We would use matched pairs, take differences in survival times for each pair, and use T =(x − µd)/(s/

√n) ∼ tn−1 where x and s are for the differences.

9.16 (a) Need to assume underlying normality for both groups, and that they have the same populationvariance.(b) Let 1 = male author. s2

p = (17 × 1.9972 + 17 × 1.652)/(18 + 18 − 2) = 3.355. H0 : µ1 = µ2 vs

Ha : µ1 6= µ2. Test statistic is T = (9.11 − 6.55)/√

3.355/18 + 3.355/18 = 2.56/0.611 = 4.19. Thet34 ≈ (t30 + t40)/2 = (1.697 + 1.684)/2 = 1.6905, so the rejection region is values < −1.6905 and> 1.6905. The test statistic is greater than 1.6905, so we reject H0 at 10% level, i.e. we concludethat gender and appearance of an author do affect the way a reader will feel about the article.(c) 90% CI for µ1−µ2 with t34 ≈ 1.6905 (from (b)) is (9.11−6.55)±1.6905×

√

3.355/18 + 3.355/18 =2.56 ± 1.032 = (1.528, 3.592). This is an approximation as we do not have the value for t34.(d) You could use a Mann-Whitney U test. A matched pairs t-test would not work as people wouldknow it was the same article.

9.17 Need to do a t-test as n2 < 30, s2p = (43×1.4682+6×2.1612)/(44+7−2) = 2.463. H0 : µ1 = µ2 vs

Ha : µ1 < µ2. Test statistic is T = (1.901−3.95)/√

2.463/44 + 2.463/7 = −2.049/0.6386 = −3.208.t49 ≈ (t40 + t60)/2 = (1.684 + 1.671)/2 = 1.6775, so the rejection region is values < −1.6775. Thetest statistic is less than -1.6775, so we reject H0 at 5% level, i.e. we conclude the new habitat isbetter than the old habitat.

9.18 (a) 1 = females : x1 = 1923.438, s1 = 383.3481, n1 = 16. 2 = males : x2 = 2090.909, s2 = 341.3709,n2 = 11. s2

p = (15 × 383.34812 + 10 × 341.37092)/(16 + 11 − 2) = 134787.1.

(b) H0 : µ1 = µ2 vs Ha : µ1 6= µ2. Test statistic is T = (1923.4−2090.9)/√

134787.1(1/16+ 1/11) =−167.5/143.7970 = −1.165. The t25 rejection region is values < −2.060 and > 2.060. The teststatistic is not in the rejection region, so we do not reject H0 at 5% level, i.e. there could be nodifference in mean weight of male and female possums.(c) H0 : population 1 is identical to population 2 vs Ha : populations 1 and 2 are not identical.

96

Ordering and ranking the two groups we get:

Females 1000 1400 1650 1750 1800 1800Males 1700 1750 1800 1850 1850rank F 1 2 3 5.5 8 8rank M 4 5.5 8 10.5 10.5

Females 1900 1925 1950 2000 2000 2200 2200 2250 2300Males 1925 2275rank F 12 13.5 15 16.5 16.5 18.5 18.5 20 22.5rank M 13.5 21

Females 2650Males 2300 2350 2550 2650rank F 26.5rank M 22.5 24 25 26.5

Summing the ranks: S1 = 207 and S2 = 171. U1 = 207 − 16×17

2= 207 − 136 = 71 and U2 =

171 − 11×12

2= 161 − 66 = 105. The test statistic is U = min(71, 105) = 71. The rejection region

is values ≤ 47, the test statistic is not in the rejection region so we do not reject H0 that thepopulations are identical, i.e. we conclude that there is no difference between weights of male andfemale possums.(d) Females: LQ = 1775, median= 1937.5 and UQ = 2200. Males: LQ = 1800, median= 1925 andUQ = 2350. The boxplots appear asymmetric so the normality required by the t-test is unlikelyto hold. The variances are probably not the same as boxplots have very different spread. TheMann-Whitney U -test seems more appropriate.

Fem

ales

Mal

es

1000 1500 2000 2500

Side by Side Boxplots of Possum Weights

weight (grams)

9.19 H0 : µ1 = µ2 vs Ha : µ1 > µ2 where 1 = control group. s2p = (11×2.0652 +11×1.7822)/(12+12−

2) = 3.72. Test statistic is T = (28.58 − 27.42)/√

3.72/12 + 3.72/12 = 1.473. The t22 rejectionregion is values > 1.321, so reject H0.

9.20 H0 : σ21

= σ22

vs Ha : σ216= σ2

2. F = 0.872/0.812 = 1.154 which is F (7, 9). α = 0.10 and this is

a 2-sided test, so the rejection region from the 5% table is values > 3.6767. We cannot reject H0

and conclude there is no evidence to suggest a difference in the two variances.

9.21 H0 : σ21

= σ22

vs Ha : σ216= σ2

2. F = 1.72/0.62 = 8.028 which is F (9, 12). The rejection region from

the 2 1

2% table is values > 3.4358. The test statistic is in the rejection region so we reject H0. This

difference in variances means we doubt our results in Exercise 9.10.

9.22 H0 : σ21 = σ2

2 vs Ha : σ21 6= σ2

2 . F = 0.732/0.72 = 1.087 which is F (13, 11). The rejection regionfrom the 2 1

2% table is values > 3.4296 (use F (12, 11) as F (13, 11) not in tables). We cannot reject

97

H0 and conclude there is no evidence to suggest a difference in the two variances.

9.23 H0 : σ21

= σ22

vs Ha : σ216= σ2

2. F = 0.6932/0.4762 = 2.12 which is F (14, 14) ≈ F (15, 14). The

rejection region from the 2 1

2% table is values > 2.9493. The test statistic is not in the rejection

region so we can’t reject H0 and conclude there is no evidence to suggest a difference in the twovariances.

9.24 H0 : σ21 = σ2

2 vs Ha : σ21 > σ2

2 . F = 383.34812/341.37092 = 1.261 which is F (15, 10). The rejectionregion from the 5% table is values > 2.8450. The test statistic is not in the rejection region so wecan’t reject H0 and conclude there is no evidence to suggest a difference in the two variances.

9.25 (a) The data are distributed asymmetrically.

Tes

t IT

est I

I

90 95 100 105 110 115 120

Side by Side Boxplots of Psychological Test Scores

weight (grams)

(b) H0 : MI = MII vs Ha : MI 6= MII.

Test I 89 89 90 91 101 105 112Test II 89 94 99 99 104 113 121rank I 2 2 4 5 9 11 12rank II 2 6 7.5 7.5 10 13 14

Summing the ranks: SI = 45 and SII = 60. UI = 45 − 7×8

2= 45 − 28 = 17 and UII = 60 − 7×8

2=

60 − 28 = 32. The test statistic is U = min(17, 32) = 17. The rejection region is values ≤ 8, thetest statistic is not in the rejection region so we do not reject H0 that the populations are identical.(c) This would make no difference as they would still be ranked 13th and 14th respectively.(d) The differences between Test I scores and Test II scores: −12, 0,−9, 6,−14,−3,−10, with xd =−6 and sd = 7.234. H0 : µd = 0 vs Ha : µd 6= 0. Test statistic is T = −6/(7.234/

√7) = −2.194.

The t6 rejection region is values < −2.447 and > 2.447, the test statistic is not in the rejectionregion, so we can’t reject H0. Alternatively, a sign test has X = 1 out of six (the zero is deletedfrom the sample), with a p-value = 2 × P (X ≤ 1) = 2 × 0.1094 > 5% confirming the conclusion ofthe t-test.

9.26 (a) H0 : M1 = M2 vs Ha : M1 6= M2.

Group 1 7 8 9 11 12 13 13 20Group 2 3 4 5 6 6 7 8 10 11 12rank 1 6.5 8.5 10 12.5 14.5 16 17 18rank 2 1 2 3 4.5 4.5 6.5 8.5 11 12.5 14.5

Summing the ranks: S1 = 103 and S2 = 68. U1 = 103 − 8 × 9/2 = 103 − 36 = 67 and U2 =68 − 10 × 11/2 = 68 − 55 = 13. The test statistic is U = min(67, 13) = 13. The rejection regionis values ≤ 17, the test statistic is in the rejection region so we reject H0 that the populations are

98

the same.(b) This would make no difference as it would still be ranked 18th and so the test statistic is notchanged.

9.27 H0 : M1 = M2 vs Ha : M1 6= M2.

Variety A 85.2 85.6 85.7 87.4 89.1Variety B 82.9 86 87.7 87.8 88.5rank A 2 3 4 6 10rank B 1 5 7 8 9

Variety A 90.3 91.5 91.6 92.2Variety B 89.8 94.0 96.1 100.2 102.0rank A 12 13 14 15rank B 11 16 17 18 19

Summing the ranks: SA = 79 and SB = 111. UA = 79 − 9 × 10/2 = 79 − 45 = 34 and UB =111 − 10 × 11/2 = 111 − 55 = 56. The test statistic is U = min(34, 56) = 34. The rejection regionis values ≤ 20, the test statistic is not in the rejection region so we do not reject H0 that thepopulations are identical.

9.28 1 = safety articles group and 2 = control group. H0 : M1 = M2 vs Ha : M1 < M2.

Safety 4 7 10 11 17 18 19 21Control 6 9 12 13 13 16rank 1 1 3 5 6 11 12 13 14rank 2 2 4 7 8.5 8.5 10

Summing the ranks: S1 = 65 and S2 = 40. U1 = 65−8×9/2 = 65−36 = 29 and U2 = 40−6×7/2 =40 − 21 = 19. The test statistic is taken from the lower population in the alternative hypothesis:U = U1 = 29. The rejection region is values ≤ 8, the test statistic is not in the rejection region sowe do not reject H0 that the populations are identical.

9.29 1 = Lower Hutt and 2 = Waitakare. H0 : M1 = M2 vs Ha : M1 6= M2.

Lower Hutt 4 7 9 12 13 13Waitakare 10 13 14 14 16 21rank 1 1 2 3 5 7 7rank 2 4 7 9.5 9.5 11 12

Summing the ranks: S1 = 25 and S2 = 53. U1 = 25 − 6 × 7/2 = 4 and U2 = 53 − 6 × 7/2 = 32.The test statistic is U = min(4, 32) = 4. Rejection region is values ≤ 5 so reject H0.

9.30 Let 1 = Original Location. The Boat Bay Location ranks are 28, 31, 32.5, 43, 45, 49, 51. S2 = 279.5and U2 = 279.5 − 7 × 8/2 = 279.5 − 28 = 251.5. Similarly S1 = 1046.5 and U1 = 1046.5 −44 × 45/2 = 1046.5 − 990 = 56.5. The test statistic is U = min(251.5, 56.5) = 56.5. Z =(U − n1n2

2)/

√

n1n2(n1 + n2 + 1)/12 = (56.5− 44×7

2)/

√

44 × 7 × (44 + 7 + 1)/12 = −97.5/36.533 =−2.6688. At the 1% level of significance, rejection region is values < −2.5758 so we reject H0. Thenormal approximation may not be very good as n2 < 25.

99

9.31 (a)

Wel

lingt

onA

uckl

and

500 1000 1500

Side by Side Boxplots of Residential Apartment Prices

price ($000)

(b) Let 1 = Wellington. H0 : M1 = M2 vs Ha : M1 < M2.

Wellington 290 370 385 460Auckland 92 100 255 310 360 360rank 1 4 8 9 10rank 2 1 2 3 5 6.5 6.5

Wellington 475 493 570 600 655 1700Auckland 490 498 670.5 1700rank 1 11 13 15 16 17 19.5rank 2 12 14 18 19.5

Summing the ranks: S1 = 122.5 and S2 = 87.5. U1 = 122.5 − 10 × 11/2 = 122.5 − 55 = 67.5 andU2 = 87.5 − 10 × 11/2 = 87.5 − 55 = 32.5. The test statistic is the U from the lower populationwhich is Wellington, so U = U1 = 67.5. The rejection region is values ≤ 23, the test statistic is notin the rejection region so we do not reject H0 that the populations are identical at the 5% level ofsignificance.

9.32 Let 1 = Pasture and 2 = Replanted forest. The sum of the ranks are: S1 = 656 and S2 = 1424.U1 = 656 − (32 × 33)/2 = 656 − 528 = 128 and U2 = 1424− (32 × 33)/2 = 1424− 528 = 896. Thetest statistic is U = min(128, 896) = 128. The rejection region is values ≤ 321, the test statistic isin the rejection region so we reject H0 in favour of differing populations. Using the normal approx-imation Z = (U − n1n2

2)/

√

n1n2(n1 + n2 + 1)/12 = (128− 32×32

2)/

√

32 × 32 × (32 + 32 + 1)/12 =−384/74.476 = −5.156. The rejection region is values < −2.5758 at the 1% level of significance,so we reject H0 using the normal approximation.

9.33 The differences between low and high altitudes (low − high) are: 9, 5, 4, 4, 11, 7 with xd = 6.667and sd = 2.875. H0 : µd = 0 vs Ha : µd > 0. Test statistic is T = 6.667/(2.875/

√6) = 5.68. The t5

rejection region is values > 2.015, the test statistic is in the rejection region, so we reject H0 andsupport the hypothesis that number of trees per quadrat does decrease.

9.34 Differences (unprotected − protected) are: 1, 2, 2, 2, 0, −2, 2, 1, 3, 1 with x = 1.2 and s = 1.398.H0 : µd = 0 vs Ha : µd > 0. Test statistic is T = 1.2/(1.398/

√10) = 2.71 which is t9. Rejection

region is values > 1.383 so reject H0.

9.35 The differences between arrivals in (2001 − 2002) are: −224,−445,−826, 310,−234 with xd =−283.8 and sd = 411.8922. H0 : µd = 0 vs Ha : µd > 0. Test statistic is T = −283.8/(411.8922/

√5) =

−1.541. The t4 rejection region is values > 2.776, the test statistic is not in the rejection region,so we can’t reject H0, i.e. there is no evidence that the terrorist attacks reduced visitor numbers.In fact, the numbers increased.

100

9.36 (a) The samples are not independent, they are two measurements of the same gecko.(b) The differences between Vol1 and Vol2 (Vol1 − Vol2) are: 0.018, 0.003, 0.021, 0.020, 0.029,−0.009, 0.027, −0.006, 0.001, 0.019, −0.003, 0.080, 0.001, 0.003, 0.006, 0.001, 0.052, 0.020, −0.005,−0.007 with xd = 0.014 and sd = 0.022. H0 : µd = 0 vs Ha : µd > 0. Test statistic is T =0.014/(0.022/

√20) = 2.846. The t19 rejection region is values > 1.729, the test statistic is in the

rejection region, so we reject H0, i.e. there is evidence that the animals settle to a lower rate ofoxygen in the second test.(c) H0 : M = 0 vs Ha : M > 0. n = 20. Comparing with m0 = 0 we have 5 below and 15above m0. Test statistic is X = 15. If H0 true X ∼ Bin(20, 0.5) and p-value of this result isP (X ≥ 15) = 0.0207 < 5% so we reject H0.

9.37 The difference between wives and husbands (wives − husbands) are: −7, −2, 13, 3, 1, 1, −3, 11,15, 15, −6, 25, 5, 13, 3. H0 : M = 0 vs Ha : M 6= 0. n = 15. Comparing with m0 = 0, there are4 below and 11 above m0. Test statistic is X = min(4, 11) = 4. If H0 true X ∼ Bin(15, 0.5) andp-value of this result is 2×P (X ≤ 4) = 2×0.0592 = 0.1184 > 5% so we do not reject H0, i.e. thereis no evidence of a difference between wives’ and husbands’ ratings.

9.38 (a) The difference between Questions D and F (D − F) are: 2, −2, 0, −1, 0, 0, 2, −1, −3, 0.H0 : M = 0 vs Ha : M 6= 0. Discarding the data where the differences are equal to m0 = 0,n = 6 and there are 4 below and 2 above m0. Test statistic is X = min(4, 2) = 2. If H0 trueX ∼ Bin(6, 0.5) and p-value of this result is 2 × P (X ≤ 2) = 2 × 0.3437 = 0.6874 > 5% so we donot reject H0, i.e. there is no difference between the scores for questions D and F.(b) The samples are not independent, they are two measurements taken from the same person.

9.39 (a) The difference between NZSX10 and ASX (NZSX10 − ASX) are: −0.06, −0.25, −0.06, −0.01,0.10, 0.04, −0.11, −0.04, −0.18, −0.15, −0.06, −0.14, 0.12, 0.08, −0.05 with xd = −0.0513 andsd = 0.1059. H0 : µd = 0 vs Ha : µd < 0. Test statistic is T = −0.0513/(0.1059/

√15) = −1.876.

The t14 rejection region is values < −1.761, the test statistic is in the rejection region, so we rejectH0, i.e. New Zealand investors did get significantly lower returns than investors in Australia.(b) H0 : M = 0 vs Ha : M < 0. n = 15. Comparing with m0 = 0 there are 11 below and4 above m0. Test statistic is X = 4. If H0 true X ∼ Bin(15, 0.5) and p-value of this result isP (X ≤ 4) = 0.0592 > 5% so we do not reject H0, i.e. there is no difference in returns.(c) The sample is symmetric, with no outliers, so the t-test should be valid.

−0.2 −0.1 0.0 0.1

Boxplot of Differences in Monthly NZSX and ASX Returns

return (%)

9.40 (a) H0 : p1 = p2 against Ha : p1 < p2 where 1 = non-smokers. p∗ = 219/327 = 0.67, p1 = 0.62,p2 = 0.77. Standard error is

√

0.67 × 0.33 × (1/218 + 1/109) = 0.055. Test statistic is Z =(0.62 − 0.77)/0.055 = −2.73. Rejection region is values < −2.3263. Z lies in the rejection regionso reject H0.(b) p-value = P (Z < −2.73) = 0.5 − 0.4968 = 0.0032.(c) 95% confidence interval is (0.62 − 0.77) ± 1.96

√

0.62 × 0.38/218 + 0.77 × 0.23/109 = −0.15 ±0.102 = (−0.252,−0.048).

9.41 H0 : p1 = p2 against Ha : p1 > p2 where 1 = fatalities in 1992. p∗ = 382/1049 = 0.364,p1 = 0.423, p2 = 0.270. Standard error is

√

0.364 × 0.636× (1/645 + 1/404) = 0.0305. Teststatistic is Z = (0.423− 0.270)/0.0305 = 5.016. Rejection region is values > 1.6449. Z lies in therejection region so reject H0, i.e. there has been a significant decrease in the proportion of fatalitiesthat involve drink driving.

9.42 H0 : p1 = p2 against Ha : p1 > p2 where 1 = group shown kittens. p∗ = 60/100 = 0.60,p1 = 37/50 = 0.74, p2 = 23/50 = 0.46. Standard error is

√

0.6 × 0.4 × (1/50 + 1/50) = 0.098.

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Test statistic is Z = (0.74 − 0.46)/0.098 = 2.86. Rejection region is values > 1.6449. Z lies in therejection region so reject H0 at 5% level.

9.43 H0 : p1 = p2 against Ha : p1 6= p2 where 1 = households in New Brighton. p∗ = 16/482 = 0.0332,p1 = 10/212 = 0.0472, p2 = 6/270 = 0.0222. Standard error is

√

0.0332× 0.9668(1/212 + 1/270) =0.0164. Test statistic is Z = (0.0472−0.0222)/0.0164 = 1.524. Rejection region is values < −1.6449and > 1.6449. Z does not lie in the rejection region so we cannot reject H0 at 10% level.

9.44 (a) p∗ = 144/700 = 0.206.(b) p1 = 0.2, p2 = 0.22 where 1 = Australia. H0 : p1 = p2 against Ha : p1 < p2. Standard erroris

√

0.206× 0.794 × (1/500 + 1/200) = 0.034. Test statistic is Z = (0.2 − 0.22)/0.034 = −0.59.Rejection region is values < −1.645. Z does not lie in rejection region so we do not reject H0.

9.45 (a) H0 : pF = pM against Ha : pF 6= pM . pF = 0.304, pM = 0.216 and p∗ = 416/1896 = 0.22. Stan-dard error is

√

0.22 × 0.78(1/79 + 1/1817) = 0.0476. Test statistic is Z = (0.304− 0.216)/0.0476 =1.85. Rejection region is values < −1.6449 and > 1.6449. Z lies in the rejection region so we rejectH0 at 10% level.(b) We rejected H0 in (a), so there is a difference in proportions, and we have evidence at the 10%significance level to fear a woman more than a man in a dark alley.

9.46 p1 = 0.066, p2 = 0.052 where 1 = Maori. and p∗ = (947 × 0.066 + 3568× 0.052)/(947 + 3568) =0.0549. H0 : p1 = p2 against Ha : p1 > p2. Standard error is

√

0.0549× 0.9451(1/947 + 1/3568) =0.0083. Test statistic is Z = (0.066 − 0.052)/0.0083 = 1.687. Rejection region is values > 2.3263.Z does not lie in the rejection region so we cannot reject H0 at 1% level of significance, i.e. we donot have enough evidence to show burglary follows the overall pattern.

9.47 (a) H0 : p1 = p2 against Ha : p1 6= p2, where 1 = Northland. p1 = 52/61 = 0.852, p2 = 115/121 =0.950 and p∗ = 167/182 = 0.918. Standard error is

√

0.918 × 0.082 × (1/61 + 1/121) = 0.0431.Test statistic is Z = (0.852 − 0.950)/0.0431 = −2.274. At the 5% level of significance, the rejectionregion is values < −1.96 and > 1.96 and the test statistic lies in the rejection region so we rejectH0.(b) Both n1 and n2 are larger than 30, so the test should be valid.

9.48 H0 : p1 = p2 against Ha : p1 < p2, where 1 = babies delivered in 2002 . p1 = 0.186, p2 = 0.199and p∗ = 259/1345 = 0.193. Standard error is

√

0.193× 0.807× (1/667 + 1/678) = 0.0215. Teststatistic is Z = (0.186 − 0.199)/0.0215 = −0.605. At the 5% level of significance, the rejectionregion is values < −1.6449, the test statistic does not lie in the rejection region so we cannot rejectH0, i.e. the increase in births by caesarian were not significant.

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Chapter 10

Exercises

The first set of problems are for the ANOVA test.

10.1 In an investigation into the literacy of young adolescent males one critical variable is thought to bewhether or not the young males have any hearing damage. A sample of fifty Year 11 male studentshad their hearing tested and then undertook a literacy test with maximum score = 50. The resultsare shown in the table below:

Hearing

Very damaged Below average Normal

12 22 17 20 26 3115 17 21 28 3115 19 22 28 3115 19 22 29 3215 19 23 29 3216 19 23 29 3217 20 24 29 3220 20 26 29 3221 20 27 30 3322 20 30 36

ni 11 19 20 n = 50Ti 190 398 609 T = 1197xi 17.27 20.95 30.45 ¯x = 23.94si 3.41 2.72 2.21

(a) Draw three side-by-side boxplots of this data.(b) List the assumptions underlying analysis of variance. Do the boxplots in (a) support these

assumptions or not?(c) Draw up an analysis of variance table for this data and test whether the observed differences in

literacy scores are significant at the 5% level. State the null and alternative hypotheses clearly.(d) Find 95% confidence intervals for the mean literacy scores for each of these three groups.

Display on a number line and comment on what they suggest.(e) What is the best point estimate of the literacy score for Year 11 males with below average

hearing? Why? What would be your estimate of the mean literacy score for Year 11 femaleswith below average hearing and why?

10.2 Groups of New Zealand Year 12 students were randomly selected from four location types: ruralareas, rural towns, provincial towns and cities and scored on a vocational aptitude test. The scoresare as follows:

Rural Rural towns Provincial towns Cities

4 30 31 1912 41 49 6644 13 22 659 26 56 4617 19 89

ni 5 4 5 5xi 17.2 27.5 35.4 57Ti 86 110 177 285

(a) Draw up an ANOVA table for this data.(b) Test whether the observed differences in the average vocational aptitude test scores are signif-

icant at the 5% level of significance. State the null and alternative hypotheses clearly definingany terms you use.

103

(c) Find 95% half confidence intervals for the mean vocational aptitude test scores for each of thesefour groups and display on a number line. Comment on what this indicates.

(d) Find a 95% confidence interval for the difference between the mean vocational aptitude testscores for the groups from provincial towns and cities. Test the claim that there is no differencein the mean vocational aptitude test scores for these two groups.

(e) What are the best point estimates for the mean vocational aptitude test score for Year 12students in rural areas and also for female Year 12 students from cities? Justify your answer.

(f) Suggest two factors, other than type of location, which might affect these scores.

10.3 There is considerable controversy surrounding the merits of various techniques for teaching readingin New Zealand. Three groups of new entrant students to Year 1 who all had English as a firstlanguage and came from the same socio-economic background were taught in one of three ways –whole word, phonics, and combination of whole word and phonics – over a six-month period withthe same total instruction time in each case. After six months each student was given a simplereading test and scored for comprehension. The marks on this test (/100) are given below:

phonics whole word combination

59 75 9478 69 8967 83 8062 81 8883 7276 79

90

n1 = 6 n2 = 7 n3 = 4x1 = 70.83 x2 = 78.43 x3 = 87.75s1 = 9.58 s2 = 7.11 s3 = 5.80

(a) Display side-by-side dotplots of this data. Based on this display does it appear that the nullhypothesis is true? Why or why not?

(b) Fill in the missing entries in the ANOVA table below:

Source Df Sum of squares Mean square

Between groups 689.6436Within groups

Total 16 1552.9412

(c) Use the table in (b) to test whether the observed differences in reading progress are significantat the 5% level. State the null and alternative hypotheses clearly, defining any terms you use.

(d) Find 95% half confidence intervals for the mean reading score in each of the three groups.Display on a number line. Comment.

(e) Find a 95% confidence interval for the difference between the mean reading score for childrentaught by the phonics and whole word methods. Test the claim that there is no difference inthe mean reading score for these two groups.

(f) What is the best point estimate of the mean reading score for children taught by the combi-nation method? Justify your answer.

(g) Draw up a table of the residuals of this data set. Use your residuals to check the assumptionof underlying normality. What can you conclude about the validity of your ANOVA?

10.4 New Zealand native frogs live a long time (up to about 30 years), and continue to grow throughouttheir lives. In 1984/5, 100 frogs (Leiopelma pakeka) were translocated from two study grids onMaud Island to a new habitat at Boat Bay. The weight gains (over the 12 years until 1997) of thefrogs shifted were compared with those of the frogs left behind, to see whether the new habitatwas as good as, or better than, the old.

The following data set shows the weight gain (in grams, over the 12 years) of frogs found and

104

weighed both in 1984/5 and 1997. Fourteen frogs had been left on Grid 1, 30 on Grid 2, and sevenhad been moved to Boat Bay in 1984/5.

Maud Island 1.67 0.25 0.43 2.33 2.85 2.22 1.07 0.10 0.85 2.20 x1 = 1.646Grid 1 0.50 2.72 1.65 4.20 s1 = 1.187

Maud Island 1.27 2.93 0.80 0.58 -0.30 3.68 -1.70 1.66 3.50 1.27 x2 = 2.02Grid 2 1.48 2.20 5.97 4.18 2.93 2.40 2.55 3.15 0.00 4.10 s2 = 1.587

1.10 3.07 3.15 1.20 1.80 3.25 1.50 1.97 1.18 -0.27

Boat Bay 2.55 3.80 2.30 4.45 8.50 3.60 2.45 x3 = 3.95s3 = 2.161

(a) Draw a line graph of the means of these groups. What does it suggest?(b) Perform an analysis of variance on this data to test whether or not the mean weight gain is

the same, irrespective of location.(c) Find a 95% confidence interval for the difference in mean weight gain between the Boat Bay

frogs and those from Grid 2, using an approximate value for the t-score. Why does the t-scorehave to be approximate?

(d) Is there any evidence that the mean weight gains in Boat Bay and Grid 2 are significantlydifferent? Explain your answer.

(Data from Bell, Pledger and Dewhurst (2004). The fate of a population of the endemic frogLeiopelma pakeka (Anura: Leiopelmatidae) translocated to restored habitat on Maud Island, NewZealand. New Zealand Journal of Zoology 31:123–131.)

10.5 The data on the New Zealand spotted skink in Exercise 3.18 seem to roughly fulfill the assumptionsbehind an analysis of variance and while the data is not continuous, we can treat it as if it werecontinuous but recorded discretely, if we convert it to average count per day. Then, a sample of 66observations from the data becomes:

Pasture 0.4 0.3 0.0 0.2 0.2 0.1 0.4 0.1 0.2 0.5 0.61.1 0.3 0.1 0.1 0.4 0.8 0.5 1.4 0.6 0.8 0.6

Replanted forest 1.5 2.4 3.1 0.8 0.4 1.8 1.4 3.3 1.1 1.6 2.61.8 0.6 1.2 1.6 1.1 0.8 1.3 1.2 1.1 0.8 0.5

Tussock 1.4 2.3 1.5 1.4 0.5 1.6 1.0 1.6 1.4 1.0 1.70.7 1.2 2.9 1.0 1.1 1.1 1.0 1.0 0.6 1.3 1.2

(a) Do an analysis of variance on this data to test whether the population mean numbers of skinkscaught per day is the same for each of the three habitats.

(b) Construct a 95% confidence interval for the difference between the mean count per day for thegroup in pasture and that in replanted forest.

10.6 Data from the baby tuatara first referred to in Exercise 2.6 is shown below. Remember that theeggs were incubated in the lab at 18, 21 or 22 degrees Celsius. After hatching, the young tuatarahad their development monitored. The measurement is SVL = snout-vent length (mm) at age 10months.

18◦C 80 82 81 82 73 85 81 83 80 8221◦C 91 84 83 90 83 87 89 85 82 8422◦C 81 83 84 87 76 84 88 83 91 84

(a) Do an analysis of variance on this data to test whether or not the egg incubation temperaturehas an effect on the mean SVL at age 10 months. Use a 5% level of significance.

(b) Calculate the residual for each observation and plot them in a boxplot. What can you concludeabout the validity of the ANOVA?

10.7 The Kapiti Island Marine Reserve, was established in 1992. Within the reserve there is no fishing,while outside it, the legal minimum length for fishing blue cod is 330mm. The data below werecollected from four marine sites in October 2003, and are the lengths (mm) of blue cod (Paraperciscolias). Sites 1 and 3 are within the reserve, and the other two are outside.

105

Site 1 in reserve 285 315 360 290 420 310 290 340280 270 255 290 300 325 280

Site 2 305 340 310 320 270 300

Site 3 in reserve 300 380 380 350 335 335 370 280320 405 300 380

Site 4 280 260 295 280 270 305 350

(a) Draw up a boxplot for each site and present them side by side. What do the plots suggestabout the four sites and the suitability of using ANOVA on the data?

(b) Perform an analysis of variance on the data at the 1% level of significance. What is yourconclusion?

(c) Give the best estimate for the mean blue cod length in Site 3. Justify your answer.

(Data collected by Carl Struthers.)

10.8 It has been suggested that the length of sentence a convicted criminal receives is related to ethnicity.In an investigation of this claim, data was collected on 44 offenders. The offenders were all male,aged between 20 and 25, and the data relates to their second burglary conviction. The lengths oftheir sentences (in months) recorded according to the ethnicity of the offender, are:

Group Data n x s

European 12 18 12 12 12 18 18 18 36 16 16.5 6.0018 12 18 12 18 18 12

Pacific Islands 18 24 16 24 20 36 30 24 18 9 23.33 6.4031

Maori 24 24 36 36 24 36 44 24 36 13 32.0 6.928236 36 24 36

Other 18 12 16 18 24 12 6 16.67 4.5019

(a) Display the four group means on a graph and join with straight lines. What does this graphsuggest?

(b) The following ANOVA table was drawn up:

Source df SS MS

Between groupsWithin groups 1545.33

Total 43 3505.00

Calculate the missing entries.(c) Test whether the observed differences in average length of sentence are significant at the 5%

level of significance.(d) Find 90% confidence intervals for the mean length of sentence for each of these four groups.(e) Construct a 95% confidence interval for the difference in mean sentence length for the Pacific

Island and Maori groups. Test the claim that there is no difference in mean sentence lengthfor these two groups.

10.9 The following data represents the areas of a particular moss (in cm2) in randomly selected smallgrids on four sites on the south east coast of Australia. Two sites (A and B) are exposed to thewind while the other two are sheltered. Site D, while sheltered, is drier than site C.

Site A Site B Site C Site D

7 6 8 72 4 4 44 6 5 2

5

x 4.3 5.3 5.6 4.5

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(a) State the null and alternative hypotheses needed in an analysis of variance of this data.(b) Complete the following ANOVA table:

Source df SS MS

Between groupsWithin groups 37.00

Total 40.92

(c) Use the table in (b) to test, at a 1% level of significance, whether the mean area of this mossis the same at all four sites.

10.10 The following data represents water temperature (◦C) of samples taken at five different locationsabove and below the waste outlet from a wood processing plant into a river. Five observations arerecorded at each location.

Group Data x

A: 5km above outlet 7 7 15 11 9 9.8B: at outlet 19 25 22 19 23 21.6C: 5km below outlet 14 18 18 19 19 17.6D: 10km below outlet 12 17 12 18 18 15.4E: 25km below outlet 11 7 10 11 15 10.8

Perform an analysis of variance on this data stating carefully your hypotheses and conclusion. Testat the 5% level of significance.

The second set of exercises are for the Kruskal-Wallis test.

10.11 In the study of a nutritional disorder that causes death if left untreated, autopsy results can give usthe liver weight of victims expressed as a percentage of body weight. The values of this percentagefor deceased sufferers in four ethnic groups are as follows:

Group A Group B Group C Group D

3.42 3.17 3.34 3.643.96 3.63 3.72 3.933.87 3.38 3.81 3.774.19 3.47 3.66 4.183.58 3.39 3.55 4.213.37 3.41 3.51 3.883.84 3.55 3.96

3.44 3.91

Test for ethnic group differences using the Kruskal-Wallis test and α = 0.01.

10.12 Using a subsample of the raw skink data of Exercise 3.18, the counts of skinks per trap totalledover a ten-day period in each habitat are as follows:

Pasture 4 3 0 2 2 1 14 12 5 0 11 5 6 8 6

Replanted forest 15 24 31 8 4 18 14 3311 16 20 1 17 12 27 26

Tussock 14 23 15 14 5 16 10 1614 10 7 10 8 12 19 17

Test for differences in skink numbers for the three habitats with a Kruskal-Wallis test, using a 5%level of significance.

107

10.13 Test the null hypothesis that three different brands of electric jugs have equal median life againstthe appropriate alternative, where the lives of several jugs randomly selected, were found to be asbelow (lifetimes are in months and all brands have a 5-year guarantee):

Brand A Brand B Brand C

73 84 8264 80 7967 81 7162 77 7570

10.14 The table below gives sweat chloride (mEq/L) for carriers of the cystic fibrosis gene and fornon-carrier controls.

Male carrier 66 22 9 38 17 42 94 48 77 31 56Female carrier 18 70 41 91 54 12 31 24 57Male control 25 15 69 57 31 48 52Female control 48 35 89 33 11 67 19 29

Test the hypothesis that the median sweat chloride concentration is the same, irrespective of genderand whether or not one carries the cystic fibrosis gene, using a Kruskal-Wallis test.

10.15 Using the data from Exercise 10.3, retest using a Kruskal-Wallis test.

10.16 Using the data in Exercise 10.8, retest whether median length of sentence is the same regardlessof ethnicity, using the Kruskal-Wallis test. Is the Kruskal-Wallis test likely to be more reliable thanANOVA in this case, and why?

10.17 Using the data from Exercise 10.9, retest using a Kruskal-Wallis test.

Solutions

10.1 (a)

very

dam

aged

Bel

ow a

vera

geN

orm

al

15 20 25 30 35

Side by Side Boxplots for Literacy levels

test score (out of 50)

(b) Assumptions: random samples, equal population variances and underlying normality of thedata. The boxplot for normal hearing looks symmetric, but the boxplots for the other groups lookasymmetric, so we may doubt the normality assumption for these two groups. Spread looks roughlythe same.(c)

108

Source df SS MS FBetween groups 2 1506.74 753.37 103.5113Within groups 47 342.08 7.278

Total 49 1848.82

H0 : all three population group means are equal, vs Ha : at least one is different. The test statisticis F = 103.5113 and 5% rejection region is values > F (2, 47) ≈ (F (2, 40) + F (2, 60))/2 = 3.191.We clearly reject H0, concluding at least one mean literacy score is different.(d) sp =

√MSE =

√7.278 = 2.698. t47 ≈ (t40+t60)/2 = 2.0105 so we will use this in the confidence

intervals as t47 is not in the tables. A 95% CI is of the form xi ± 2.0105 × 2.698/√

ni. For verydamaged hearing, we get 17.27±2.0105×2.698/

√11 = 17.27±1.63 = (15.64, 18.90), below average

hearing gives 20.95±1.244 = (19.71, 22.19) and normal hearing gives 30.45±1.2125 = (29.24, 31.66).The confidence intervals overlap somewhat, but the normal hearing do get higher scores than theother two groups. Half-confidence intervals show separation between normal and others.

Confidence intervals for the individual population means

mu

Very damaged

Below average

Normal

5 10 15 20 25 30 35 40

(e) The best point estimate of the mean is the sample estimate x2 = 20.95 as we rejected H0 inpart (c). This data is for males only, so probably no estimate is sensible. Using the assumptionthat males and females are the same, you could use 20.95.

10.2 (a)Source df SS MS F

Between groups 3 4227.42 1409.14 4.066Within groups 15 5199 346.6

Total 18 9426.42

(b) H0 : all four population group means are equal, vs Ha : at least one is different. The test statis-tic is F = 4.066 and rejection region is values > F (3, 15) = 3.2874. We reject H0, and conclude atleast one group mean is different.(c) sp =

√MSE =

√346.6 = 18.617. A half confidence interval is of the form xi±18.617/

√ni. Ru-

ral: 17.2±18.495/√

5 = 17.2±8.326 = (8.874, 25.526), rural towns: 27.5±9.309 = (18.191, 36.809),provincial towns: 35.4 ± 8.326 = (27.074, 43.726), and cities: 57 ± 8.326 = (48.674, 65.326). Thereis no common value of µ which passes through all intervals, which is consistent with the result ofthe ANOVA test. The average for cities is much higher than the other group means, and theremay be a difference between rural scores and those for provincial towns.

Confidence intervals for the individual population means

mu

Rural

Rural towns

Provincial towns

Cities

0 10 20 30 40 50 60 70

(d) (35.4− 57)± 2.131× 18.617√

1/5 + 1/5 = −21.6± 25.0913 = (−46.691, 3.491). H0 : the meansfrom provincial towns and cities are the same vs Ha : they are different. 0 is in the the 95%confidence interval so we cannot reject H0.(e) The best point estimates are the sample means, i.e. x = 17.2 for rural areas. For females inthe city we have no specific data. Assuming they are similar to the males, we could use the males’average for the cities x = 57.

109

(f) Examples include gender, ethnicity and socio-economic status.

10.3 (a) Based on the dotplots, it appears that the combination group have higher scores than the othergroups, and the phonics scores are slightly lower.

60 65 70 75 80 85 90 95

Pho

nics

Com

bo

Reading Test Scores for Three Groups With Different Training

Test score (/20)

(b)Source df SS MS F

Between groups 2 689.6436 344.8218 5.592Within groups 14 863.2976 61.6641

Total 16 1552.941

(c) H0 : all three population group means are equal, vs Ha : at least one is different. The teststatistic is F = 5.592. The rejection region is values > F (2, 14) = 3.7389, so we reject H0.(d) 95% half CIs all have form xi ± sp/

√ni. For the phonics group the CI is (67.62, 74.04), for

whole word it is (75.46, 81.4) and for combination group it is (83.82, 91.68). Each CI is clearlyseparated from the others with no overlap so we can conclude that all the µi differ.(e) (−16.97, 1.77). Zero lies inside this CI so cannot reject that µ1 = µ2.(f) H0 was rejected so the best point estimate is x3 = 87.75.(g) The residuals are xij − xi: −11.83, 7.17, −3.83, −8.83, 12.17, 5.17, −3.43, −9.43, 4.57, 2.57,−6.43, 0.57, 11.57, 6.25, 1.25, −7.75, 0.25. min = −11.83, LQ = −6.43, median = 0.57, UQ = 5.17and max = 12.17. IQR = 5.17 − −6.43 = 11.6, LQ − 1.5IQR = −6.43 − 17.4 = −23.83 andUQ + 1.5IQR = 5.17 + 17.4 = 22.57, there are no values less than −23.83 or more than 22.57, sothere are no outliers. The boxplot is as follows:

−10 −5 0 5 10

Boxplot of Residuals

Residual

The median differs slightly from the mean (0). The boxplot is fairly symmetric, so the ANOVA islikely to be valid.

10.4 (a) The graph suggests that the Boat Bay mean is higher than the means for the Maud Islandsites.

110

Average weight gain for frogs at three locations

Location

Mea

n w

eigh

t gai

n

2.0

2.5

3.0

3.5

4.0

Maud Island, Grid 1 Maud Island, Grid 2 Boat Bay

(b)Source df SS MS F

Between groups 2 26.6945 13.347 5.367Within groups 48 119.3799 2.487

Total 50 146.0745

H0 : all three population group means are equal, vs Ha : at least one is different. The test statisticis F = 5.367 and rejection region is values > F (2, 48) ≈ F (2, 40) = 3.2317 at the 5% level. Wereject H0, the means are not all equal, i.e. mean weight gain does depend on the location of thefrogs.(c) Using t48 ≈ (t40 + t60)/2 = 2.0105 and sp =

√2.487 = 1.577, 95% confidence interval is

(3.95− 2.17)± 2.0105× 1.577√

1/7 + 1/30 = 1.78± 1.33 = (0.45, 3.11). The t-score is approximateas there is no value for t48 in the tables.(d) 0 is not in the confidence interval, so this is evidence that the difference between the two is notzero, i.e. there is a significant difference.

10.5 (a)Source df SS MS F

Between groups 2 13.075 6.538 18.931Within groups 63 21.757 0.345

Total 65 34.832

H0 : all three population group means are equal, vs Ha : at least one is different. The test statisticis F = 18.931 and rejection region is values > F (2, 65) ≈ F (2, 60) = 3.1504. We reject H0, themean number of skinks caught each day is not the same in all locations.(b) For pasture x = 0.441, for replanted forest x = 1.455. Using t63 ≈ t60 = 2.000 and sp =√

0.345 = 0.587, 95% confidence interval is (0.441−1.455)±2.000×0.587√

1/22 + 1/22 = −1.014±0.354 = (−1.368,−0.660).

10.6 (a)Source df SS MS F

Between groups 2 123.8 61.9 5.043Within groups 27 331.4 12.274

Total 29 455.2

H0 : all three population group means are equal, vs Ha : at least one is different. The test statisticis F = 5.043. Rejection region is values > 3.3541 (F (2, 27) with α = 0.05). So reject H0.(b) Residuals for 18 ◦C are: −0.9, 1.1, 0.1, 1.1, −7.9, 4.1, 0.1, 2.1, −0.9, 1.1. Residuals for 21 ◦Care: 5.2, −1.8, −2.8, 4.2, −2.8, 1.2, 3.2, −0.8, −3.8, −1.8. Residuals for 22 ◦C are: −3.1, −1.1,−0.1, 2.9, −8.1, −0.1, 3.9, −1.1, 6.9, −0.1. These have median = −0.1, LQ = −1.8, UQ = 2.1,min = −8.1 and max = 6.9, with a very symmetric boxplot. The normality assumption is probably

111

OK.

10.7 (a) Site 1: min = 255, LQ = 280 median = 290, UQ = 325 and max = 420. Site 2: min = 270,LQ = 300 median = 307.5, UQ = 320 and max = 340. Site 3: min = 280, LQ = 310 median= 342.5, UQ = 380 and max = 405. Site 4: min = 260, LQ = 270 median = 280, UQ = 305 andmax = 350.

Site

1S

ite 2

Site

3S

ite 4

250 300 350 400

Side by Side Boxplots for Blue Cod Lengths

length (mm)

Sites 2 and 3 are fairly symmetric but the other 2 samples are asymmetric. The ranges and IQRsare quite different, so we may doubt that the equal variance assumption. This ANOVA may notbe valid.

Source df SS MS FBetween groups 3 15589 5196.33 3.815Within groups 36 49039 1362.194

Total 39 16951.19

H0 : all four population group means are equal vs Ha : at least one is different. The test statistic isF = 3.815. Rejection region is values > F (3, 36) ≈ (F (3, 30) + F (3, 40))/2 = (4.5097 + 4.3126)/2 =4.4112 at 1% level of significance. We do not reject H0, and conclude all 4 population means couldbe identical.(c) As we accepted H0, the best point estimate for the mean blue cod length at site 3 is the grandmean ¯x = 315.75.

10.8 (a) The graph suggests a marked difference in sentence length for Maori which is greater thanPacific Islands which is in turn greater than the remaining 2 groups.

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Average length of sentence for second burglary convictions

Location

Mea

n le

ngth

of s

ente

nce

2025

30

European Pacific Islands Maori Other

(b)Source df SS MS F

Between groups 3 1959.67 653.223 16.91Within groups 40 1545.33 38.633

Total 43 3505.00

(c) H0 : all four population group means are equal vs Ha : at least one is different. The test statisticis F = 16.91. Rejection region is values > F (3, 40) = 2.8387 at 5% level of significance. We rejectH0, and conclude the observed differences are significant at the 5% level.(d) sp =

√MSE =

√38.633 = 6.216 and t40 = 1.684. The 90% CIs are of the form xi ±

1.684 × 6.216/√

ni. European: 16.5 ± 2.617 = (13.883, 19.117), Pacific Islands: 23.33 ± 3.489 =(19.841, 26.819), Maori: 32± 2.903 = (29.097, 34.903) and Other: 16.67± 4.273 = (12.397, 20.943).(e) (23.33 − 32.00) ± 2.021 × 6.216 ×

√

1/9 + 1/13 = −8.67 ± 5.447 = (−14.117,−3.223). 0 is notin the 95% confidence interval so we would reject H0 : that the means are the same between thetwo groups.

10.9 (a) H0 : all four population group means are equal vs Ha : at least one is different.(b)

Source df SS MS FBetween groups 3 3.92 1.307 0.318Within groups 9 37.00 4.111

Total 12 40.92

(c) The test statistic is F = 0.318. Rejection region is values > F (3, 9) = 6.9919 at 1% level ofsignificance. The test statistic is not in the rejection region so we do not reject H0, so the meanarea of moss could well be the same at all sites.

10.10 H0 : all five population group means are equal vs Ha : at least one is different.

Source df SS MS FBetween groups 4 475.66 118.92 14.745Within groups 20 161.30 8.065

Total 24 636.96

The test statistic is F = 14.745. Rejection region is values > F (4, 20) = 2.866 at 5% level ofsignificance. We reject H0, and conclude a significant difference in average water temperatures atthe five sites.

10.11 H0 : all four populations have the same median percentage vs Ha : at least one median differs.Group A has ranks: 7, 25.5, 21, 28, 13, 3, 20 with RA = 117.5. Group B has ranks: 1, 14, 4,9, 5, 6, 11.5, 8 with RB = 58.5. Group C has ranks: 2, 17, 19, 16, 11.5, 10 with RC = 75.5.

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Group D has ranks: 15, 24, 18, 27, 29, 22, 25.5, 23 with RD = 183.5.∑

R2

i /ni = 7559.154 andK = (12 × 7559.154)/(29× 30) − 3 × 30 = 104.264 − 90 = 14.26. The rejection region is values> 11.345 (χ2

3at 1% level), so reject H0.

10.12 H0 : all three populations have the same median number of skinks vs Ha : at least one mediandiffers. Sample 1 has ranks: 10.5, 9, 1.5, 7, 7, 4, 30, 4, 7, 13, 1.5, 24.5, 13, 15.5, 19, 15.5 withR1 = 182. Sample 2 has ranks: 33.5, 44, 47, 19, 10.5, 40, 30, 48, 24.5, 36, 42, 4, 38.5, 26.5, 46,45 with R2 = 534.5. Sample 3 has ranks: 30, 43, 33.5, 30, 13, 36, 22, 36, 30, 22, 17, 22, 19, 26.5,41, 38.5 with R3 = 459.5.

∑

R2

i /ni = 33122.16 and K = (12 × 33122.156)/(48× 49) − 3 × 49 =168.991− 147 = 21.991. The rejection region is values > 5.991 (χ2

2 at 5% level), so reject H0.

10.13 H0 : all three brands have same median lifetimes vs Ha : at least one median differs. Sample 1 hasranks: 6, 2, 3, 1, 4 with R1 = 16. Sample 2 has ranks: 13, 10, 11, 8 with R2 = 42. Sample 3 hasranks: 12, 9, 5, 7 with R3 = 33.

∑

R2

i /ni = 764.45 and K = (12×764.45)/(13× 14)−3×14 = 8.403.The rejection region is values > 5.991 (χ2

2at 5% level), so reject H0.

10.14 H0 : all four populations have the same median sweat chloride concentration vs Ha : at least onemedian differs. Sample 1 has ranks: 28, 8, 1, 17, 5, 19, 35, 21, 32, 13, 25 with R1 = 204. Sample2 has ranks: 6, 31, 18, 34, 24, 3, 13, 9, 26.5 with R2 = 164.5. Sample 3 has ranks: 10, 4, 30,26.5, 13, 21, 23 with R3 = 127.5. Sample 4 has ranks: 21, 16, 33, 15, 2, 29, 7, 11 with R4 = 134.∑

R2

i /ni = 11356.785 and K = (12 × 11356.785)/(35× 36) − 3 × 36 = 108.1599 − 108 = 0.1599.The rejection region is values > 7.815 (χ2

3 at 5% level), so we do not reject H0.

10.15 H0 : all three reading systems have same median lifetime score (if tested on very large groups ofchildren) vs Ha : at least one median differs. Sample 1 has ranks: 1, 8, 3, 2, 12.5, 7 with R1 = 33.5.Sample 2 has ranks: 6, 4, 12.5, 11, 5, 9, 16 with R2 = 63.5. Sample 3 has ranks: 17, 15, 10, 14 withR3 = 56.

∑

R2

i /ni = 1547.08 and K = (12 × 1547.08)/(17× 18) − 3 × 18 = 6.67. The rejectionregion is values > 5.991 (χ2

2 at 5% level), so reject H0.

10.16 H0 : all four populations have the same median length of sentence vs Ha : at least one mediandiffers. Sample 1 has ranks: 5, 17.5, 5, 5, 5, 17.5, 17.5, 17.5, 39, 17.5, 5, 17.5, 5, 17.5, 17.5, 5 withR1 = 214. Sample 2 has ranks: 17.5, 29, 10.5, 29, 24, 39, 34, 29, 17.5 with R2 = 229.5. Sample 3 hasranks: 29, 29, 39, 39, 29, 39, 44, 29, 39, 39, 39, 29, 39 with R3 = 462. Sample 4 has ranks: 17.5, 5,10.5, 17.5, 29, 5 with R4 = 84.5.

∑

R2

i /ni = 26323.31 and K = (12×25650.19)/(44× 45)−3×45 =159.525− 135 = 24.535. The rejection region is values > 7.815 (χ2

3at 5% level), so we reject H0.

By looking at the boxplots, the distributions are asymmetric and so we doubt normality for eachpopulation, and the Kruskal-Wallis test will be more reliable than ANOVA.

Eur

opea

nP

acifi

c Is

land

Mao

riO

ther

15 20 25 30 35 40 45

Boxplot of Length of Sentence

Length of sentence (months)

10.17 H0 : all four groups have the same median area of moss vs Ha : at least one median differs. SampleA has ranks: 11.5, 1.5, 4.5 with RA = 17.5. Sample B has ranks: 9.5, 4.5, 9.5 with RB = 23.5.

114

Sample C has ranks: 13, 4.5, 7.5 with RC = 25. Sample D has ranks: 11.5, 4.5, 1.5, 7.5 withRD = 25.

∑

R2i /ni = 650.749 and K = (12×650.749)/(13× 14)−3×14 = 42.9065−42 = 0.9065.

The rejection region is values > 7.815 (χ23

at 5% level), so we do not reject H0.

115

Chapter 11

Exercises

This set of exercises is for one-way chi-square, and goodness of fit tests.

11.1 If a die is fair you would expect that one-sixth of all throws would result in a “1”, one-sixth in a“2” etc. In 100 throws of a die the following results occurred:

1 2 3 4 5 6

Frequency 19 12 11 16 20 22

Using a goodness of fit test with α = 0.05, test whether or not the die is fair.

11.2 A certain plant comes in four varieties, which according to a genetic model theoretically occur inthe ratio 9:3:3:1. In 245 seeds of this plant it was observed that the numbers in the four varietieswere 153, 52, 28 and 12 respectively. Does this data support the frequencies predicted by thegenetic model? Test at the 5% level of significance.

11.3 In 1994, 17% of New Zealanders used EFTPOS on a daily basis, while 75% used it at least once aweek, and 8% had never used it. A random sample of 1000 people conducted in 2001 included 168individuals who used EFTPOS daily, 752 who used it at least once a week, and only 80 who hadnever used it. Test whether or not the set of proportions changed significantly between 1994 and2001.(Data from the AC Nielsen Consumer Finance Monitor.)

11.4 In a field study on finding tuatara at three sites on Stephen’s Island, the number of sites at whichtuatara were found on any one visit to the island followed a binomial distribution with n = 3 andp = 0.4. In 24 trips to the island the number of trips in which no tuatara were found at all wasseven. For eight trips tuatara were found at one site only, while six trips found tuatara at two sitesand three trips found tuatara at all three sites.

(a) Complete the table below and perform a goodness of fit test on this data using α = 0.05. LetX =number of sites at which tuatara were found.

X Probability Number observed Number expected

0 0.216 71 0.432 82 0.288 63 0.064 3

(b) Is there evidence that a binomial distribution may not apply, in particular that p = 0.4 is notconstant over time? Give reasons for your answer.

11.5 The supply of many services to visitors to New Zealand is evenly distributed across the 12 monthsof the year. The following data are the actual visitor arrivals in 2002.

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec7421 6998 7280 4902 3978 3950 4600 4268 4480 5600 6300 7400

Test whether the arrivals are significantly different across months at the 5% level of significance.What might cause the arrivals to differ?(Data thanks to David Lowrie, Statistics New Zealand.)

11.6 Ambulance call-outs over one week (in a particular town) were as follows:

Monday Tuesday Wednesday Thursday Friday Saturday Sunday125 83 92 86 109 148 134

116

Does this data support the contention that ambulance call-out frequency is the same irrespectiveof the day of the week? Use α = 0.05.

11.7 Of the 200 winners in one week in a scratch card competition, 112 had even-numbered cards and88 had odd-numbered cards. Is this significantly different from 50:50? Use α = 0.10.

11.8 It is often claimed that basically 39% New Zealand voters are National voters, 42% are Labourvoters and the rest split across all other parties. In a random sample of 500 Dunedin voters 230said they would vote National and 250 Labour. Does this data support the claim at a 5% level ofsignificance?

The following problems use the contingency table analysis to examine for relationships between categor-ical variables.

11.9 The table below gives data on three types of major offences by gang affiliation of sentenced inmatesin New Zealand jails.

Gang Affiliation

Offence Member Associate None Total

Violence 215 200 1485 1900Sexual violence 46 45 1009 1100Drug dealing 39 25 326 390

Total 300 270 2820 3390

(a) Present this data in a stacked bar chart.(b) What assumptions are needed before you can be confident of drawing valid conclusions from

the table? Are they likely to be satisfied in this case? Give a reason for your answer.(c) It is wished to test whether the data provides sufficient evidence to indicate a relationship

between the type of major crime an inmate has been convicted of and their gang affiliation,assuming your assumptions in (b) hold.

i. State the null and alternative hypotheses that you would use.ii. Assuming your null hypothesis is correct, give a table of expected frequencies corresponding

to the above table of observed frequencies.iii. Test whether or not the gang affiliation of a convicted inmate is independent of the type

of major crime convicted of at a 5% level of significance.

(Data based on the census of prison inmates, 1999, Department of Corrections.)

11.10 A survey of 800 New Zealanders in 2001 recorded the numbers who had been victims of violentcrime at least once in the previous year. The results are shown in the table below:

European Maori Pacific Other Total

Victim of violent crime 51 25 6 2 84Not a victim 559 100 34 23 716

Total 610 125 40 25 800

(a) What assumptions do you need before being confident of drawing valid conclusions from thetable? Are they likely to be satisfied in this case? Give reasons for your answer.

(b) Test whether the data provides sufficient evidence to indicate a relationship between being avictim or not, and ethnicity, using a 5% level of significance.

(Data based on New Zealand National Survey of Crime Victims. 2001. Ministry of Justice.)

11.11 In an investigation into the distribution of particular native trees, three areas of the South Islandwere targeted. A quadrat at a particular altitude was selected randomly in each of three nationalparks and the numbers of ribbonwood and lacebark trees (including saplings and seedlings) werecounted within each quadrat. The results were:

117

Nelson Lakes Mount Aspiring Mount Cook

Ribbonwood 24 17 23Lacebark 4 13 9

(a) It is wished to test whether the data provide sufficient evidence to indicate a relationshipbetween the type of tree and location. Compute a test statistic for this data based on observedand expected frequencies, and test whether the type of tree is independent of location at a 1%level of significance.

(b) Originally a fourth location in Central Otago had been investigated with three ribbonwoodsand two lacebarks counted. Explain why this location was not included in the analysis.

11.12 The number of new non-residential building authorisations per year in three North Island regionsis shown in the table below:

Year Taranaki Auckland Wellington

1997 241 690 1791999 215 783 1972001 199 831 244

(a) Present this data in a stacked bar chart.(b) Test whether or not the number of new residential building authorisations in a year is inde-

pendent of region at a 1% level of significance.

11.13 As part of a local study on perception of environmental issues, 300 18- to 25-year-olds living inWellington gave their opinion on whether the motorway bypass through the centre city should goahead. The subjects were classified according to their principal occupational group and resultswere:

Full-time work Carer forStudent inc. apprentices children at home Unemployed

Yes to bypass 32 60 29 12No to bypass 68 42 41 16

Test whether the data provides sufficient evidence to indicate a relationship between principaloccupation and views on whether the bypass should be built. Are your assumptions satisfied?State your hypotheses, test statistic, rejection region and conclusion clearly.

11.14 The Department of Conservation is looking to change its policy on the logging of beech forest.A survey of 400 people were given a draft of the proposed new policy and asked for their views,which are summarised below:

Age of respondents

Opinion 18 - 29 30 - 39 40 - 59 60+

Agree 29 77 2 54Oppose 61 115 6 56

(a) Why is it necessary to amalgamate some columns of this table in order to do a valid test ofwhether there is a relationship between age and attitude to the proposed policy?

(b) Amalgamating the 30 - 39 and 40 - 59 categories, we have:

Opinion 18 - 29 30 - 59 60+

Agree 29 79 54Oppose 61 121 56

Does the data provide sufficient evidence to indicate that attitude to the proposed policy isindependent of the age of respondents at a 5% level of significance? At 1%?

118

11.15 Fear of flying, while relatively common, can be very inconvenient. A study looking at whetherthere was a relationship between age and fear of flying resulted in the following data obtained from240 males:

Fear of Flying

Age group Always Sometimes Never

Under 25 8 7 4026 - 40 57 47 20Over 40 25 16 20

Test whether degree of fear of flying is independent of age group at a 5% level of significance. Whatfurther information would you need in order to be confident of your conclusion?

11.16 Two hundred former psychiatric patients who had been hospitalised for depression took part ina study designed to lift their overall mood. In particular patterns of waking were assessed. Thestudy subjects were put into one of three treatment programmes; a physical fitness programme, amusic and art therapy programme, and the third group was given no special therapy. At the end ofa 4-month trial the participants were monitored for two weeks and among other things questionedabout their frequency of early morning wakening (a common symptom of depression). The resultswere:

Frequency of early morning awakening

Almost nightly 2 - 5 nights/week Once/week or not at all

Physical fitness 24 16 30Music and art 17 39 4No programme 23 31 16

Calculate row and column totals for this contingency table and use them to calculate expectedfrequencies for each cell assuming no dependence between treatment programme and frequencyof early morning awakening. Test whether the data provide sufficient evidence to indicate a re-lationship between treatment programme and frequency of early morning awakening. Test usingα = 0.01.

11.17 The following data were collected on customer satisfaction at a Wellington store:

Not satisfied Satisfied Very satisfied

Male 13 42 32Female 23 30 15

(a) Do these data suggest that the gender of the customer, and their satisfaction level are related?Use a 5% level for the test.

(b) What recommendation would you give to the manager of the store?

11.18 The table below displays deaths in the 15- to 24-year-old males in New Zealand in 1990 by ethnicityand cause of death.

Cause of death Maori Non-Maori

Vehicle crashes 33 234Suicide 10 101Drowning 2 15Assault 2 11Other 13 89

(a) Test whether or not cause of death and ethnicity are independent in this age group of malesusing a 5% level of significance.

119

(b) Should the rows for drowning and assault be amalgamated since they have rather small num-bers, especially for Maori? Justify your answer.

(c) Draw up a table similar to the one above using percentage of Maori and non-Maori in eachcause of death instead of numbers. Why is this helpful?

(Mortality and Demographic Data, Department of Health, 1993 NZHIS.)

11.19 As part of a study on whether the willingness of people to offer assistance to a person in distresswas influenced by the number of bystanders, 60 subjects observed a person who appeared to behaving an asthma attack. Twenty subjects were alone (apart from the person having the attack)while the others had various numbers of other people around. These bystanders were part of theexperiment and offered no help. The results were:

No bystanders One bystander ≥ 2 bystanders

Help offered 14 7 4No help offered 6 13 16

Test whether or not help is offered is independent of the number of other bystanders at a 5% levelof significance.

11.20 Commuters who come into Wellington to work come from a variety of residential areas. Thefollowing data are for 1976 and 1991:

Year Kapiti Coast Wairarapa Upper Hutt Lower Hutt Porirua

1976 2060 200 3521 9697 72681991 2421 480 3360 11613 7137

Are the home regions of commuters into Wellington independent of year, i.e. has the pattern ofcommuting changed over time? Test at the 5% level of significance.(Data thanks to Brad Patterson, published in McConchie, Winchester and Willis, Dynamic Welling-ton, 2000, Institute of Geography, VUW.)

11.21 Delegates to the last three ICME conferences come from all over the world. The table below showsthe venue of the last three conferences and the country of origin of some of the delegates.

Conference Venue

Delegates from Denmark Japan Spain Total

Nordic nations 388 108 126 622Canada & USA 369 239 327 935UK & Ireland 114 92 208 414Australia & New Zealand 104 77 146 327China, Japan & SE Asia 155 1090 150 1395Northern Europe 136 83 114 333Southern Europe 147 79 1027 1253

Total 1413 1768 2098 5279

(a) Assuming that the numbers attending each conference are independent of the country of originof the delegates, draw up a table of expected frequencies.

(b) What are the degrees of freedom associated with this contingency table? Find the rejectionregion for a chi-square statistic with those degrees of freedom using a 1% level of significance.

(c) Test the assumption in (a) at the 1% level of significance. You will not need to evaluate all theterms of your test statistic. Why?

11.22 A survey was conducted among a random sample of 6,672 adults and one of the questions asked waswhether the respondent was right-handed, ambidextrous or left-handed. The following frequencytable was obtained where “other” includes both left-handed and ambidextrous:

120

Men Women

Right-handed 2,780 3,281Other 311 300

(a) Test whether gender and handedness are independent at the 5% level of significance using Yates’continuity correction in your test statistic and stating clearly your hypotheses and conclusion.

(b) The total number of ambidextrous men and women in the sample was four. Explain whyleft-handedness and ambidextrous were combined.

11.23 Referring back to Exercise 9.17, there were 44 Maud Island frogs that were not relocated to BoatBay. These frogs were in one of two locations: Grid 1 or Grid 2. The weight gains for these frogsover the 12-year period were as follows:

Grid 1 1.67 0.25 0.43 2.33 2.85 2.22 1.07 0.10 0.85 2.200.50 2.72 1.65 4.20

Grid 2 1.27 2.93 0.80 0.58 -0.30 3.68 -1.70 1.66 3.50 0.004.10 1.10 3.07 3.15 1.20 1.80 3.25 1.50 1.97 1.181.27 1.48 2.20 5.97 4.18 2.93 2.40 2.55 3.15 -0.27

Count the numbers of frogs in each location that made a weight gain of less than 2.33g and thosewhose gain was 2.33g or more and use this to complete the following table:

Grid 1 Grid 2

Weight gain < 2.33gWeight gain ≥ 2.33g

Test whether the weight gain of the frogs is independent of their location on Maud Island. Use a5% level of significance.

11.24 A random sample of 30 females and 30 males was collected and their romantic attachments studied.All subjects were students. Of interest was the level of anxiety people feel about their romanticattachments. Subjects respond to items such as: “I rarely worry about my partner leaving me” ona Likert scale and results were collated into an overall anxiety score. Results were as follows:

Males Females

Below average score 24 20Average and above score 6 10

Test whether level of anxiety about romantic attachments is independent of gender at a 5% levelof significance.(Data thanks to Chris Sibley.)

11.25 In a clinical trial of cannabis for the relief of pain in severe chronic arthritis, 400 patients receivedtreatment with cannabis oil and a control group of 400 received treatment with a similar-lookingplacebo. After a six-month trial results were:

Placebo Cannabis

Improved 238 144Not improved 162 256

Conduct a chi-square test to decide if these results give evidence at the 5% level of significancethat the cannabis oil is effective in reducing pain in arthritis.

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Solutions

11.1 H0 : pi = 1/6 vs Ha : at least one differs. Expected frequency in each category 100 × 1/6 = 16.67.Test statistic χ2 = (19− 16.67)2/16.67 + · · ·+ (22− 16.67)2/16.67 = 5.959. The rejection region isvalues > 11.070 (χ2

5at 5%). Cannot reject H0, the die seems to be fair.

11.2 H0 : p1 = 9/16, p2 = 3/16, p3 = 3/16, p4 = 1/16 vs Ha : at least one differs from that stated. Theobserved frequencies are 153, 52, 28 and 12. The expected frequencies are 9

16× 245 = 137.8125,

3

16× 245 = 45.9375, 3

16× 245 = 45.9375 and 1

16× 245 = 15.3125 respectively. The test statis-

tic is χ2 = (153 − 137.8125)2/137.8125 + (52 − 45.9375)2/45.9375 + (28 − 45.9375)2/45.9375 +(12 − 15.3125)2/15.3125 = 1.674 + 0.800 + 7.004 + 0.717 = 10.195. The rejection region is val-ues > 7.815, so we reject H0 at the 5% level. The data does not support the frequencies predictedby the genetic model.

11.3 H0 : p1 = 0.17, p2 = 0.75, p3 = 0.08 vs Ha : at least one differs. The observed frequencies are 168,752 and 80. The expected frequencies are 0.17×1000 = 170, 0.75×1000 = 750 and 0.08×1000 = 80respectively. The test statistic is χ2 = (168 − 170)2/170 + (752 − 750)2/750 + (80 − 80)2/80 =0.024 + 0.005 + 0 = 0.029. The rejection region is values > 5.991 at the 5% level, so we do notreject H0, i.e. the proportions have not changed significantly between 1994 and 2001.

11.4 (a) H0 : X ∼ Bin(3, 0.4), vs Ha : X is not binomial with p = 0.4. The expected frequenciesare 0.216 × 24 = 5.184, 0.432 × 24 = 10.368, 0.288 × 24 = 6.912, 0.064 × 24 = 1.536. Only 75%have expected value > 5, so we will combine the last 2 groups. (for the new group 3: observed= 9, expected = 8.448). The test statistic is χ2 = (7 − 5.184)2/5.184 + (8 − 10.368)2/10.368 +(9 − 8.448)2/8.448 = 0.6362+ 0.5408+ 0.03607 = 1.2131. The rejection region is values > 5.991 atthe 5% level, so we do not reject H0, and conclude that the binomial distribution is an appropriateone for the data.(b) The data follows the binomial distribution at each trip, but over time this distribution may bechanging. It is noticeable that in each trip you would expect 3 × 0.4 = 1.2 tuatara so in 24 tripsyou expect about 24 × 1.2 = 28.8 whereas only 25 were observed.

11.5 H0 : pi = 1/12 vs Ha : at least one pi 6= 1/6. Total number of visitor arrivals = 67177. If these areevenly distributed, the expected values are = 1/12 × 67177 = 5598.083. The test statistic is χ2 =(7421− 5598.083)2/5598.083+ · · ·+ (7400− 5598.083)2/5598.083 = 593.6 + · · ·+ 580.00 = 3874.9.The χ2

11 rejection region is values > 19.675, the test statistic is in rejection region, so visitor arrivalnumbers are not evenly distributed throughout the year. Note that you only need to work out thefirst term which is already > 19.675.

11.6 H0 : pi = 1/7 vs Ha : at least one pi 6= 1/6. Total number of call-outs = 777. If the frequency isindependent of the day of the week, the expected values are = 1/7× 777 = 111. The test statisticis χ2 = (125 − 111)2/111+ · · ·+(134 − 111)2/111 = 1.766+7.063+3.252+5.631+0.036+12.333+4.766 = 34.85 The rejection region is values > 12.592, the test statistic is in rejection region, so wereject H0, i.e. ambulance call-out frequency does depend on what day of the week it is.

11.7 H0 : pi = 0.5 vs Ha : pi 6= 0.5. Expected frequencies are 200 × 0.5 = 100. Test statisticχ2 = (112 − 100)2/100 + (88 − 100)2/100 = 2.88. The rejection region is values > 2.706 (χ2

1at

10%). Reject H0. Notice that this is equivalent to a large sample test of a proportion, which givesa p-value of 8.97% and the same conclusion.

11.8 H0 : pℓ = 0.42, pn = 0.39, po = 0.19. Observed frequencies were 250, 230 and 20 with n = 500.Expected frequencies are 0.42×500 = 210, 0.39×500 = 195 and 0.19×500 = 95 respectively. Teststatistic χ2 = (250 − 210)2/210 + (230 − 195)2/195 + (20 − 95)2/95 = 7.619 + 6.282 + 59.211 =73.112. The rejection region is values > 5.991 (χ2

2 at 5%). Reject H0, the claim is false (in Dunedinat least).

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11.9 (a)

Violence Sexual violence Drug dealing

NoneAssociateMember

Major offence classified by gang affiliation

Fre

quen

cy

050

010

0015

00

(b) We need a random sample and each observation can only fit into one category. We may doubtthese assumptions in this case as people in gangs may know each other and they might havecommitted more than one offence, i.e. they are likely to share beliefs and behaviour.(c) i. H0 : there is no association between gang affiliation and offence vs Ha : there is an association.ii. Member Associate None Total

Violence 168.14 151.33 1580.53 1900Sexual Violence 97.35 87.61 915.04 1100Drug Dealing 34.51 31.06 324.43 390Total 300 270 2820 3390

iii. Test statistic is χ2 = (215 − 168.14)2/168.14+· · ·+(326 − 324.43)2/324.43 = 13.06+· · ·+0.01 =93.72. There are (r − 1) × (c − 1) = 2 × 2 = 4 degrees of freedom and χ2

4 at 5% = 9.488. The teststatistic is greater than 9.488, so we reject H0, and conclude there is an association between gangaffiliation and offence.

11.10 (a) Data must enter the table independently so we must have no relatives in the sample and noone must be counted twice under two ethnicities. Unlikely to be satisfied, as being a victim, likebeing a criminal, tends to run in families because similar sets of social circumstances tend to apply.(b) H0 : whether or not one is a victim of violent crime is independent of ethnicity vs Ha :there is a relationship between ethnicity and whether or not one is a victim of violent crime.Expected frequencies are (L-R, top to bottom): 64.05, 13.125, 4.2, 2.625, 545.95, 111.875, 35.8,22.375 and 25% of these are less than 5 so we should amalgamate. Combining Pacific and Othergroups, with observed (expected) frequencies 8 (6.825) and 57 (58.175), the test statistic is χ2 =(51 − 64.05)2/64.05 + · · · + (57 − 58.175)2/58.175 = 2.659 + · · · + 0.0237 = 15.2014. The rejectionregion is values > 5.991 (χ2

2 at 5%). The test statistic lies in the rejection region, so we reject H0.

11.11 (a) H0 : there is no association vs Ha : there is an association. Expected frequencies:

Nelson Lakes Mount Aspiring Mount Cook TotalRibbonwood 19.91 21.33 22.76 64Lacebark 8.09 8.67 9.24 26Total 28 30 32 90

Test statistic is χ2 = (24 − 19.91)2/19.91 + · · · + (9 − 9.24)2/9.24 = 0.840 + · · · + 0.006 = 5.963.There are (r−1)×(c−1) = 2×1 = 2 degrees of freedom and χ2

2 at 1% = 9.210. The test statistic isless than 9.210, so we do not reject H0, and conclude that there is no relationship between locationand type of tree.(b) The values in the cells are not big enough, you would get too many expected values < 5.

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11.12 (a)

1997 1999 2001

WellingtonAucklandTaranaki

Non−residential building authorisations by region

Fre

quen

cy

020

040

060

080

010

0012

00

(b) H0 : region and year of non-residential building authorisations are independent vs Ha : theyare dependent.

Taranaki Auckland Wellington Total1997 203.14 714.57 192.29 11101999 218.70 769.29 207.10 11952001 233.16 820.14 220.70 1274Total 655 2304 620 3579

Test statistic is χ2 = 7.055+ · · ·+2.460 = 17.217. There are (r−1)× (c−1) = 2×2 = 4 degrees offreedom and χ2

4 at 1% = 13.277. Rejection region is values > 13.277, so we reject H0 and concludethat there is an association. From the plot, we see that Auckland’s permits are increasing, whilethose in Taranaki are decreasing.

11.13 H0 : occupation and opinion are independent, vs Ha : they are not independent. Table of expectedvalues:

Student Full time Carer Unemployed TotalYes 44.33 45.22 31.03 12.41 133No 55.67 56.78 38.97 15.59 167Total 100 102 70 28 300

Test statistic is χ2 = 3.431+ · · ·+0.011 = 15.106. There are (r−1)× (c−1) = 1×3 = 3 degrees offreedom and χ2

3 at 5% = 7.815, so rejection region is values > 7.815 and we reject H0. We concludethat there is an association between principal occupation and views on the bypass.

11.14 (a) r1 = 162, r2 = 238, c1 = 90, c2 = 192, c3 = 8 and c4 = 110 with n = 400. Expected frequenciesare (L-R, top to bottom): 36.45, 77.76, 3.24, 44.55, 53.55, 114.24, 4.76, 65.45. There are two cellswith expected frequencies less than 5 (25%) so we must amalgamate.(b) The new expected frequencies are: 36.45, 81.0, 44.55, 53.55, 119.0, 65.45. H0 : age and opinionon this issue are independent vs Ha : age and opinion are associated. This has (r− 1)× (c− 1) = 2degrees of freedom. Critical value at 5% is 5.991. Test statistic is χ2 = (29− 36.45)2/36.45+ · · ·+(56 − 65.45)2/65.45 = 6.011. We reject H0. At 1% level of significance critical value = 9.21 so at1% we cannot reject H0.

11.15 H0 : Fear of flying and age are independent vs Ha : they are dependent. Table of expected values:

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Always Sometimes Never TotalUnder 25 20.63 16.04 18.33 5526 -40 46.5 36.17 41.33 124Over 40 22.88 17.79 20.33 61Total 90 70 80 240

Test statistic is χ2 = 7.728 + · · ·+ 0.005 = 55.440. There are (r − 1)× (c − 1) = 2 × 2 = 4 degreesof freedom and χ2

4at 5% = 9.488, rejection region is values > 9.488, we reject H0, and conclude

that age and fear of flying are dependent.

11.16 H0 : there is no relationship between type of treatment and frequency of early morning wakeningvs Ha : there is a relationship. Table of expected values:

Always Sometimes Never TotalPhysical fitness 22.4 30.1 17.5 70Music and Art 19.2 25.8 15.0 60No programme 22.4 30.1 17.5 70Total 64 86 50 200

Test statistic is χ2 = 0.114 + 6.605 + · · ·+ 0.129 = 30.892. There are (r − 1)× (c− 1) = 2 × 2 = 4degrees of freedom and χ2

4at 1% = 13.277, rejection region is values > 13.277, the test statistic is

in the rejection region so we reject H0, i.e. there is a relationship between treatment programmeand frequency of early morning awakening.

11.17 (a) H0 : there is no relationship between gender and satisfaction level vs Ha : there is a relationship.Table of expected values:

Not satisfied Satisfied Very satisfied TotalMale 20.21 40.41 26.38 87Female 15.79 31.59 20.62 68Total 36 72 47 155

Test statistic is χ2 = (13 − 20.21)2/20.21+ · · ·+(15 − 20.62)2/20.62 = 2.572+ · · ·+1.531 = 8.735.There are (r − 1)× (c− 1) = 2 × 1 = 2 degrees of freedom, rejection region is values > 5.991. Thetest statistic is in the rejection region so we reject H0, i.e. there is a relationship between genderand customer satisfaction.(b) 23/68 = 34% of female customers are not satisfied compared with 15% of male customers. Themanager of the store needs to find a way to make female customers feel more satisfied.

11.18 (a) H0 : Cause of death and ethnicity are independent vs Ha : they are not independent. Tableof expected values:

Maori Non-Maori TotalVehicle crashes 31.41 235.59 267Suicide 13.06 97.94 111Drowning 2.00 15.00 17Assault 1.53 11.47 13Other 12.00 90.00 102Total 60 450 510

Test statistic is χ2 = (33 − 31.41)2/31.41 + · · · + (89 − 90)2/90 = 0.080 + · · · + 0.011 = 1.161570.There are (r − 1)× (c− 1) = 1 × 4 = 4 degrees of freedom, rejection region is values > 9.488. Thetest statistic is not in the rejection region so we do not reject H0 that cause of death and ethnicityare independent.(b) We consider amalgamating rows if less than 80% have expected frequency > 5, the table in (a)shows that 8/10 = 80% have expected frequency > 5, so there is no need to amalgamate.(c) Percentage of total for each cause of death (% to 1dp.):

125

Cause of death Maori Non-MaoriVehicle crashes 55.0 52.0Suicide 16.7 22.4Drowning 3.3 3.3Assault 3.3 2.4Other 21.7 19.8

The population sizes are very different which makes comparing the numbers difficult. By convertingto percentages, we see that the two groups are similar.

11.19 H0 : whether or not help is offered is independent of the number of other bystanders vs Ha :there is a relationship between whether or not help is offered and the number of other bystanders.Expected frequencies are: 8.33, 8.33, 8.33, 11.667, 11.667, 11.667 with 2 degrees of freedom. Criticalvalue at 5% = 5.991. Test statistic is χ2 = (14−8.33)2/8.33+ · · ·+(16−11.667)2/11.667 = 10.837.So reject H0.

11.20 H0 : Patterns of commuting into Wellington is independent of year vs Ha : Patterns of commutinginto Wellington is dependent on year. Table of expected values:

Kapiti Coast Wairarapa Upper Hutt Lower Hutt Porirua Total1976 2134.24 323.87 3277.33 10149.66 6860.90 227461991 2346.76 356.13 3603.67 11160.34 7544.10 25011Total 4481 680 6881 21310 14405 47757

Test statistic is χ2 = 2.582+ · · ·+21.968 = 214.664. There are (r−1)× (c−1) = 4×1 = 4 degreesof freedom, rejection region is values > 9.488. The test statistic is in the rejection region so wereject H0, i.e. patterns of commuting have changed over time.

11.21 (a) Table of expected values:

Denmark Japan SpainNordic Nations 166.49 208.32 247.20Canada & USA 250.27 313.14 371.59UK & Ireland 110.81 138.65 164.53Australia & New Zealand 87.53 109.52 129.96China, Japan & SE Asia 373.39 467.20 554.41Northern Europe 89.13 111.53 132.34Southern Europe 335.38 419.64 497.97

(b) There are (r−1)× (c−1) = 2×6 = 12 degrees of freedom and χ212

at 1% = 26.217, so rejectionregion is values > 26.217. (c) χ2 = (388 − 166.49)2/166.49 + · · · = 294.725 + · · · > 26.217. Thefirst term of the test statistic > 26.217, and all terms in the test statistic are positive so we canreject H0.

11.22 (a) H0 : gender and handedness are independent vs Ha : gender and handedness are not inde-pendent. Expected frequencies are: 2807.94, 3253.06, 283.06, 327.94 with 1 degree of freedom.Using Yates’ correction, the numerator of the test statistic is (29.94 − 0.5)2 = 29.442 for all cells.Test statistic is χ2 = 29.442/2807.94 + 29.442/3253.06 + 29.442/283.06 + 29.442/327.94 = 5.455.Rejection region is values > 3.841, so reject H0.(b) Because these would both have expected numbers less than 5 which is 33.3% and greater thanthe allowed 20%.

11.23 Here is the completed contingency table with observed / expected frequencies:

Grid 1 Grid 2 TotalWeight gain < 2.33g 10 / 8.59 17 / 18.41 27Weight gain ≥ 2.33g 4 / 5.41 13 / 11.59 17Total 14 30 44

H0 : Weight gain of frogs on Maud Island is independent of their location vs Ha : they are notindependent. Using Yates’ correction, the numerator of the test statistic is (1.41 − 0.5)2 = 0.912

for all cells. Test statistic is χ2 = 0.912/8.59 + 0.912/5.41 + 0.912/18.41 + 0.912/11.59 = 0.096 +

126

0.153 + 0.045 + 0.071 = 0.365. The test has 1 degree of freedom, with rejection region at 5% level> 3.841. We do not reject H0 that weight gain is independent of location on Maud Island.

11.24 H0 : Romantic attachments and gender are independent vs Ha : they are not independent. Ex-pected values are: 22, 22, 8 and 8. Using Yates’ correction, the numerator of the test statisticis (2 − 0.5)2 = 1.52 for all cells. The test statistic is χ2 = 1.5222 + 1.5222 + 1.528 + 1.528 =0.102 + 0.102 + 0.281 + 0.281 = 0.767. The test has 1 degree of freedom, with rejection region at5% level > 3.841. We do not reject H0, and conclude that the level of anxiety about romanticattachments is independent of gender.

11.25 H0 : whether or not the patient improves is independent of the use of cannabis or a placebo vsHa : they are not independent. The expected frequencies are: 191, 191, 209, 209, with 1 degree offreedom. Using Yates’ correction, the numerator of the test statistic is (47 − 0.5)2 = 46.52 for allcells. The test statistic is χ2 = 46.52/191 + 46.52/191 + 46.52/209 + 46.52/209 = 43.33. Rejectionregion is values > 3.841, so reject H0.

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Chapter 12

Exercises

In each of the following exercises state carefully your hypotheses, test statistic, rejection region andconclusion.

12.1 In Exercise 4.14, test at the 5% level the hypothesis that there is no linear relationship betweenthe metabolic rates in trial 1 and trial 2 (i.e. the slope of the regression line = 0).

12.2 In Exercise 4.17 test at the 1% level whether or not the slope of the true regression line is zero.What does this imply about the nature of the relationship between social dominance scores andsupport for biculturalism in principle scores?

12.3 Continuing Exercise 4.22, test at the 1% level the hypothesis that there is no linear relationshipbetween spatial ability score and accuracy of recollection (i.e. the slope of the regression line = 0).

12.4 Using the data from Exercise 4.27, test whether or not a linear relationship can be said to existbetween median income per region and percentage of white collar workers in that region, using a1% level of significance.

12.5 Test whether the true slope of the regression line in the situation described in Exercise 4.29 is zerousing a 5% level of significance.

12.6 Using your results from Exercise 4.32, test whether or not a linear relationship exists betweenincubation temperature and snout-vent length in tuatara. Use α = 0.05.

12.7 Using the data in Exercise 4.25, the manager would like to see a profit of at least $20 for each $1spent on advertising. Test whether or not the slope exceeds 20 at the 5% level.

These exercises are for tests of the intercept.

12.8 In Exercise 4.22 test whether the true intercept of the regression line is −1.5 using a 5% level ofsignificance.

12.9 Test H0 : α = 0.04 in Exercise 4.24 using a 5% level of significance.

12.10 Using the data in Exercise 4.25, interpret the intercept term in this situation. Is the interceptsignificantly greater than 8?

12.11 Using Exercise 4.27, test at the 5% level whether or not the intercept for the regression line forthe population from which that sample was drawn is −5.

12.12 From Exercise 4.33, test H0 : α = 50 using a 5% level of significance.

12.13 The trainer in Exercise 4.13 is worried that the system favours the younger trainees over thementors. If the programme is fair to both partners, then we would expect α = 0 and β = 1.

(a) Test whether or not β = 1 at the 5% level.(b) Test whether or not α = 0 at the 5% level.(c) What conclusion can you draw about the programme from the sample data?

In the following exercises, give 95% prediction intervals in the following situations.

12.14 Using the data in Exercise 4.14, for a trial 1 metabolic rate of 0.1 mL/g.

12.15 Using the data in Exercise 4.17, for an SDO score of 2.41.

12.16 Using the data in Exercise 4.20, for a weight of strontium of 5.75%.

12.17 Using the data in Exercise 4.21, for a house with 3 bedrooms.

12.18 Using the data in Exercise 4.25, for advertising expenditure of $750.

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Solutions

12.1 H0 : β = 0 vs Ha : β 6= 0. SXX = 0.02446, SY Y = 0.01314, SXY = 0.01421. Have σ = 0.0165 andtest statistic T = 0.581/(0.0165/

√0.02446) = 5.513 and this is t18. The rejection region is values

> 2.101 and < −2.101, the test statistic is in the rejection region, so reject H0, concluding thatthe slope of the regression line is significantly different from zero.

12.2 H0 : β = 0 vs Ha : β 6= 0. SXX = 2.8065, SY Y = 3.2267, SXY = 0.5787. Have σ = 0.5574 andtest statistic T = 0.2062/(0.5574/

√2.8065) = 0.6197 and this is t10. The rejection region is values

> 3.169 and < −3.169, the test statistic is not in the rejection region, so we can’t reject H0. Thereis no linear relationship between the two scores.

12.3 H0 : β = 0 vs Ha : β 6= 0. Have σ = 3.618 and test statistic T = 7.342/(3.618/√

11.489) = 6.88and this is t8. The rejection region is values > 3.355, so reject H0.

12.4 H0 : β = 0 vs Ha : β 6= 0. SXX = 62477500, SY Y = 471.5244, SXY = 139971.2. Have σ = 3.3588and test statistic T = 0.00224/(3.3588/

√62477500) = 5.2722 and this is t14. The rejection region

is values > 2.977 and < −2.977, the test statistic is in the rejection region, so we can reject H0.There is a linear relationship between the median income per region and percentage of white collarworkers.

12.5 H0 : β = 0 vs Ha : β 6= 0. SXX = 2392.9, SY Y = 4719.6, SXY = 3032.8. Have σ = 10.46292and test statistic T = 1.26742/(10.46292/

√2392.9) = 5.9256 and this is t8. The rejection region is

values > 2.306 and < −2.306, the test statistic is in the rejection region, so we can reject H0.

12.6 H0 : β = 0 vs Ha : β 6= 0. Have σ = 4.870 and test statistic T = 1.269/(4.87/√

34.667) = 1.53 andthis is t10. The rejection region is values > 2.228 and < −2.228 so cannot reject H0.

12.7 H0 : β = 20 vs Ha : β > 20. SXX = 0.18167, SY Y = 88.3825, SXY = 3.9. Have σ = 0.6825and test statistic T = (21.46789− 20)/(0.6825/

√0.18167) = 0.9167 and this is t10. The rejection

region is values > 1.812, the test statistic is not in the rejection region, so we do not reject H0. Weconclude that the true slope does not exceed 20.

12.8 H0 : α = −1.5 vs Ha : α 6= −1.5. Again have σ = 3.618, with se(a) = 3.618 ×√

62.145/11.489 =8.415. The test statistic is (−1.48− (−1.5))/8.415 = 0.00234 and this is t8. The rejection region isvalues > 2.306 or < −2.306, so we cannot reject H0.

12.9 H0 : α = 0.04 vs Ha : α 6= 0.04. SXX = 0.0245, SY Y = 0.01314, SXY = 0.01421. σ = 0.0165, withse(a) = 0.0165 ×

√

0.0117/0.0245 = 0.01142. The test statistic is T = (0.0295− 0.04)/0.01142 =−0.919 and this is t18. The rejection region is values > 2.101 or < −2.101, the test statistic is notin the rejection region, so we cannot reject H0.

12.10 H0 : α = 8 vs Ha : α > 8. From 12.7, have σ = 0.6825, so se(a) = 0.6825 ×√

0.41625/0.1817 =1.0331. The test statistic is T = (11.12867− 8)/1.0331 = 3.0285 and this is t10. At 5% significance,the rejection region is values > 1.812, the test statistic is in the rejection region, so we can rejectH0.

12.11 H0 : α = −5 vs Ha : α 6= −5. From 12.4, have σ = 3.3588, so se(a) = 3.3588√

298456250/62477500 =7.3411. The test statistic is T = (−5.8312− (−5))/7.3411 = −0.1132 and this is t14. At 5% signif-icance, the rejection region is values > 2.145 or < −2.145, the test statistic is not in the rejectionregion, so we do not reject H0, so the intercept could be −5.

12.12 H0 : α = 50 vs Ha : α 6= 50. Have σ = 6.547, with se(a) = 6.547 ×√

550/1090 = 4.651. Thetest statistic is (71.28 − 50)/4.651 = 4.58 and this is t8. The rejection region is values > 2.306 or< −2.306, so we reject H0.

12.13 (a) Test H0 : β = 1 vs Ha : β 6= 1. SXX = 718, SY Y = 563.875, SXY = 514. Have σ = 5.7142.Test statistic is T = (0.7158 − 1)/(5.7142/

√718) = −1.333 and this is t6. The rejection region is

values > 2.447 and < −2.447. The test statistic is not in the rejection region, so we do not rejectH0.(b) H0 : α = 0 vs Ha : α 6= 0. se(a) = 5.7142 ×

√

450.75/26.7955 = 4.5275. The test statistic isT = 11.7733/4.5275 = 2.600 and this is also t6. The test statistic is in the rejection region, so we

129

also reject H0 : α = 0.(c) We reject H0 : α = 0, but not β = 1. The evidence indicates that the programme is not fair toboth partners.

12.14 The prediction is Y = 0.0294 + 0.5805 × 0.1 = 0.0875. From 12.1, we have σ = 0.0165, sose(Y ) = 0.0165×

√

1 + 1/20 + (−0.0026)2/0.02446 = 0.01688. The confidence interval is 0.0875±2.101× 0.01688 = 0.0875± 0.0355 giving (0.0521, 0.1230).

12.15 The prediction is Y = 2.86099 + 0.206189 × 2.41 = 3.3579. From 12.2, we have σ = 0.5574, sose(Y ) = 0.5574 ×

√

1 + 1/12 + (0.1192)2/2.8065 = 0.5815. The confidence interval is 3.3579 ±2.228× 0.5815 = 3.3579 ± 1.2957 giving (2.0622, 4.6536).

12.16 The prediction is Y = 6.1786 + 2.2551 × 5.75 = 19.1454. SXX = 4.59375, SY Y = 48.5547,SXY = 10.3594. We have σ = 2.0491 and se(Y ) = 2.0491×

√

1 + 1/8 + (0.9375)2/4.59375 = 2.351.The confidence interval is 19.1454± 2.447× 2.351 = 19.1454± 5.7528 giving (7.3572, 30.9336).

12.17 The prediction is Y = 74.8 + 49 × 3 = 221.8, i.e. $221,800. We have σ = 45.285 and se(Y ) =45.285 ×

√

1 + 1/24 + 0.5422/47.958 = 46.35. The confidence interval is 221.8 ± 2.074 × 46.35 =221.8± 96.1 giving (125.7, 317.9).

12.18 The prediction is Y = 11.12867 + 21.46789 × 0.75 = 27.23. From 12.7, we have σ = 0.6825, sose(Y ) = 0.6825 ×

√

1 + 1/12 + (0.1167)2/0.18167 = 0.7345. The confidence interval is 27.23 ±2.228× 0.7345 = 27.23 ± 1.64 giving (25.59, 28.87).

130